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Supersymmetry Demystified

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Supersymmetry Demystified Patrick Labelle, Ph.D.

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Copyright © 2010 by The McGraw-Hill Companies, Inc. All rights reserved. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written permission of the publisher. ISBN: 978-0-07-163642-1 MHID: 0-07-163642-0 The material in this eBook also appears in the print version of this title: ISBN: 978-0-07-163641-4, MHID: 0-07-163641-2. All trademarks are trademarks of their respective owners. Rather than put a trademark symbol after every occurrence of a trademarked name, we use names in an editorial fashion only, and to the benefit of the trademark owner, with no intention of infringement of the trademark. Where such designations appear in this book, they have been printed with initial caps. McGraw-Hill eBooks are available at special quantity discounts to use as premiums and sales promotions, or for use in corporate training programs. To contact a representative please e-mail us at [email protected]. Information contained in this work has been obtained by The McGraw-Hill Companies, Inc. (“McGraw-Hill”) from sources believed to be reliable. However, neither McGraw-Hill nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw-Hill nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw-Hill and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought. TERMS OF USE This is a copyrighted work and The McGraw-Hill Companies, Inc. (“McGrawHill”) and its licensors reserve all rights in and to the work. Use of this work is subject to these terms. Except as permitted under the Copyright Act of 1976 and the right to store and retrieve one copy of the work, you may not decompile, disassemble, reverse engineer, reproduce, modify, create derivative works based upon, transmit, distribute, disseminate, sell, publish or sublicense the work or any part of it without McGraw-Hill’s prior consent. You may use the work for your own noncommercial and personal use; any other use of the work is strictly prohibited. Your right to use the work may be terminated if you fail to comply with these terms. THE WORK IS PROVIDED “AS IS.” McGRAW-HILL AND ITS LICENSORS MAKE NO GUARANTEES OR WARRANTIES AS TO THE ACCURACY, ADEQUACY OR COMPLETENESS OF OR RESULTS TO BE OBTAINED FROM USING THE WORK, INCLUDING ANY INFORMATION THAT CAN BE ACCESSED THROUGH THE WORK VIA HYPERLINK OR OTHERWISE, AND EXPRESSLY DISCLAIM ANY WARRANTY, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO IMPLIED WARRANTIES OF MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE. McGraw-Hill and its licensors do not warrant or guarantee that the functions contained in the work will meet your requirements or that its operation will be uninterrupted or error free. Neither McGraw-Hill nor its licensors shall be liable to you or anyone else for any inaccuracy, error or omission, regardless of cause, in the work or for any damages resulting therefrom. McGraw-Hill has no responsibility for the content of any information accessed through the work. Under no circumstances shall McGraw-Hill and/or its licensors be liable for any indirect, incidental, special, punitive, consequential or similar damages that result from the use of or inability to use the work, even if any of them has been advised of the possibility of such damages. This limitation of liability shall apply to any claim or cause whatsoever whether such claim or cause arises in contract, tort or otherwise.

CONTENTS

Acknowledgments CHAPTER 1

CHAPTER 2

Introduction 1.1 What Is Supersymmetry, and Why Is It Exciting (the Short Version)? 1.2 What This Book Is and What It Is Not 1.3 Effective Field Theories, Naturalness, and the Higgs Mass 1.4 Further Reading A Crash Course on Weyl Spinors 2.1 Brief Review of the Dirac Equation and of Some Matrix Properties 2.2 Weyl versus Dirac Spinors 2.3 Helicity 2.4 Lorentz Transformations and Invariants 2.5 A First Notational Hurdle 2.6 Building More Lorentz Invariants Out of Weyl Spinors 2.7 Invariants Containing Lorentz Indices 2.8 A Useful Identity 2.9 Introducing a New Notation 2.10 Quiz

xiii 1 1 3 6 11 13 14 17 22 24 26 28 32 33 34 40

vi CHAPTER 3

CHAPTER 4

CHAPTER 5

Supersymmetry Demystified New Notation for the Components of Weyl Spinors 3.1 Building Lorentz Invariants 3.2 Index-Free Notation 3.3 Invariants Built Out of Two Left-Chiral Spinors 3.4 The Notation for ±iσ 2 3.5 Notation for the Indices of σ μ and σ¯ μ 3.6 Quiz The Physics of Weyl, Majorana, and Dirac Spinors 4.1 Charge Conjugation and Antiparticles for Dirac Spinors 4.2 CPT Invariance 4.3 A Massless Weyl Spinor 4.4 Adding a Mass: General Considerations 4.5 Adding a Mass: Dirac Spinors 4.6 The QED Lagrangian in Terms of Weyl Spinors 4.7 Adding a Mass: Majorana Spinors 4.8 Dirac Spinors and Parity 4.9 Deﬁnition of the Masses of Scalars and Spinor Fields 4.10 Adding a Mass: Weyl Spinor 4.11 Relation Between Weyl Spinors and Majorana Spinors 4.12 Quiz Building the Simplest Supersymmetric Lagrangian 5.1 Dimensional Analysis 5.2 The Transformation of the Fields 5.3 Transformation of the Lagrangian 5.4 Quiz

43 51 53 56 57 60 65

67 68 70 72 73 74 76 78 81 82 85 86 88

89 90 91 95 99

Contents

vii

CHAPTER 6

The Supersymmetric Charges and Their Algebra 101 6.1 Charges: General Discussion 102 6.2 Explicit Representations of the Charges and the Charge Algebra 107 6.3 Finding the Algebra Without the Explicit Charges 111 6.4 Example 113 6.5 The SUSY Algebra 115 6.6 Nonclosure of the Algebra for the Spinor Field 125 6.7 Introduction of an Auxiliary Field 128 6.8 Closure of the Algebra 131 6.9 Quiz 131

CHAPTER 7

Applications of the SUSY Algebra 7.1 Classiﬁcation of States Using the Algebra: Review of the Poincar e´ Group 7.2 Effects of the Supercharges on States 7.3 The Massless SUSY Multiplets 7.4 Massless SUSY Multiplets and the MSSM 7.5 Two More Important Results 7.6 The Algebra in Majorana Form 7.7 Obtaining the Charges from Symmetry Currents 7.8 Explicit Supercharges as Quantum Field Operators 7.9 The Coleman-Mandula No-Go Theorem 7.10 The Haag-Lopuszanski-Sohnius Theorem and Extended SUSY 7.11 Quiz

133

Adding Interactions: The Wess-Zumino Model 8.1 A Supersymmetric Lagrangian With Masses and Interactions 8.2 A More General Lagrangian

159

CHAPTER 8

134 140 144 146 147 150 151 153 156 157 158

160 167

Supersymmetry Demystified

viii 8.3 8.4 8.5 CHAPTER 9

The Full Wess-Zumino Lagrangian The Wess-Zumino Lagrangian in Majorana Form Quiz

Some Explicit Calculations 9.1 Refresher About Calculations of Processes in Quantum Field Theory 9.2 Propagators 9.3 One Point Function 9.4 Propagator of the B Field to One Loop 9.5 Putting It All Together 9.6 A Note on Nonrenormalization Theorems 9.7 Quiz

173 175 176 177 178 182 185 189 201 203 204

CHAPTER 10

Supersymmetric Gauge Theories 205 10.1 Free Supersymmetric Abelian Gauge Theory 206 10.2 Introduction of the Auxiliary Field 209 10.3 Review of Nonabelian Gauge Theories 211 10.4 The QCD Lagrangian in Terms of Weyl Spinors 217 10.5 Free Supersymmetric Nonabelian Gauge Theories 219 10.6 Combining an Abelian Vector Multiplet With a Chiral Multiplet 222 10.7 Eliminating the Auxiliary Fields 232 10.8 Combining a Nonabelian Gauge Multiplet With a Chiral Multiplet 233 10.9 Quiz 237

CHAPTER 11

Superspace Formalism 11.1 The Superspace Coordinates 11.2 Example of Spacetime Translations

239 240 242

Contents

ix 11.3 Supersymmetric Transformations of the Superspace Coordinates 11.4 Introduction to Superﬁelds 11.5 Aside on Grassmann Calculus 11.6 The SUSY Charges as Differential Operators 11.7 Constraints and Superﬁelds 11.8 Quiz

CHAPTER 12

CHAPTER 13

Left-Chiral Superfields 12.1 General Expansion of Left-Chiral Superﬁelds 12.2 SUSY Transformations of the Component Fields 12.3 Constructing SUSY Invariants Out of Left-Chiral Superﬁelds 12.4 Relation Between the Superpotential in Terms of Superﬁelds and the Superpotential of Chapter 8 12.5 The Free Part of the Wess-Zumino Model 12.6 Why Does It All Work? 12.7 Quiz

244 248 250 255 261 268 269 269 273 277

285 288 292 295

Supersymmetric Gauge Field Theories in the Superfield Approach 297 13.1 Abelian Gauge Invariance in the Superﬁeld Formalism 297 13.2 Explicit Interactions Between a Left-Chiral Multiplet and the Abelian Gauge Multiplet 303 13.3 Lagrangian of a Free Supersymmetric Abelian Gauge Theory in Superﬁeld Notation 306 13.4 The Abelian Field-Strength Superﬁeld in Terms of Component Fields 308 13.5 The Free Abelian Supersymmetric Lagrangian from the Superﬁeld Approach 313 13.6 Supersymmetric QED 317 13.7 Supersymmetric Nonabelian Gauge Theories 322 13.8 Quiz 324

x

Supersymmetry Demystified

CHAPTER 14

SUSY Breaking 325 14.1 Spontaneous Supersymmetry Breaking 326 14.2 F-Type SUSY Breaking 330 14.3 The O’Raifeartaigh Model 331 14.4 Mass Spectrum: General Considerations 334 14.5 Mass Spectrum of the O’Raifeartaigh Model for m 2 ≥ 2g 2 M 2 335 14.6 The Supertrace 339 14.7 D-Type SUSY Breaking 340 14.8 Second Example of D-Type SUSY Breaking 343 14.9 Explicit SUSY Breaking 344 14.10 Quiz 350

CHAPTER 15

Introduction to the Minimal Supersymmetric Standard Model 15.1 Lightning Review of the Standard Model 15.2 Spontaneous Symmetry Breaking in the Standard Model 15.3 Aside on Notation 15.4 The Left-Chiral Superﬁelds of the MSSM 15.5 The Gauge Vector Superﬁelds 15.6 The MSSM Lagrangian 15.7 The Superpotential of the MSSM 15.8 The General MSSM Superpotential 15.9 Quiz

CHAPTER 16

Some Phenomenological Implications of the MSSM 16.1 Supersymmetry Breaking in the MSSM 16.2 The Scalar Potential, Electroweak Symmetry Breaking, and All That 16.3 Finding a Minimum 16.4 The Masses of the Gauge Bosons 16.5 Masses of the Higgs

351 351 360 364 365 368 368 371 377 378

379 380 381 384 387 392

Contents

xi 16.6 Masses of the Leptons, Quarks, and Their Superpartners 16.7 Some Other Consequences of the MSSM 16.8 Coupling Uniﬁcation in Supersymmetric GUT 16.9 Quiz

399 400 401 412

Final Exam

413

References

417

APPENDIX A

Useful Identities

421

APPENDIX B

Solutions to Exercises

427

APPENDIX C

Solutions to Quizzes

461

APPENDIX D

Solutions to Final Exam

469

Index

475

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ACKNOWLEDGMENTS

My heartfelt thanks go to Judy Bass at McGraw-Hill and Preeti Longia Sinha at Glyph International for their guidance, patience, and dedication. I am indebted to Professor Kevin Cahill of the University of New Mexico for many insightful comments on an early draft. A special thanks goes to Professor Pierre Mathieu of Universit´e Laval for introducing me to supersymmetry when I was an undergraduate and for his continuing support in my research career. I am deeply grateful to my sisters, Micheline and Maureen, for being there for me throughout the years, through thick and thin. I also want to thank my three rescued cats, Sean, Blue, and Fanny, for keeping me company and keeping me entertained during countless hours of writing and proofreading. Adopt a rescued animal if you can! I would like to dedicate this book to the memory of my mother, Pierrette, and the three siblings I have lost, Monique, Johnny, and Anne. They are no longer with us in body, but they are still very present in my heart.

ABOUT THE AUTHOR

Patrick Labelle has a Ph.D. in theoretical physics from Cornell University and held a postdoctoral position at McGill University. He has been teaching physics at the college level for 12 years at Bishop’s University and Champlain Regional College in Sherbrooke, Quebec. Dr. Labelle spent summers doing research both at the Centre Europ´een de Recherches Nucl´eaires (CERN) in Geneva, home of the Large Hadron Collider, and at Fermilab, the world’s second-largest particle accelerator near Chicago. He participated in the French translation of Brian Greene’s The Elegant Universe (in the role of technical adviser).

CHAPTER 1

Introduction

1.1 What Is Supersymmetry, and Why Is It Exciting (the Short Version)? The job of a particle theorist is pretty simple. It consists of going through the following four steps: 1. Build a theory that is mathematically self-consistent. 2. Calculate physical processes using the theory. 3. Call experimentalist friends to see if the results of the theory agree with experiments. 4. If they agree, celebrate and hope that the Nobel Committee will notice you. If they don’t, go back to step 1. That’s pretty much all a particle theorist does with his or her life. However, if one randomly tries every theory that comes to one’s mind, one soon realizes that most of them lead to results that have nothing to do with the real world. It also

2

Supersymmetry Demystified

becomes clear that requiring certain speciﬁc properties and symmetries helps to weed out unphysical theories. One such requirement is that the theory must be Lorentz invariant.* Another symmetry that has proved unexpectedly powerful in the construction of the standard model of particle physics is gauge invariance. However, the standard model is widely considered to be an incomplete theory, even at energies as low as the weak scale (about 100 GeV). One key reason is the socalled hierarchy problem that comes from the fact that the mass of the Higgs particle† receives large corrections from loop diagrams (see Section 1.3 for a more detailed discussion). These corrections, in principle, can be canceled by a ﬁne-tuning of some parameters of the standard model, but this highly contrived solution seems very unnatural to most physicists (for this reason, the hierarchy problem is also often called a naturalness or ﬁne-tuning problem). There is therefore a strong impetus to build new theories going beyond the standard model (while reproducing all of its successes, of course). For this, as always, it is useful to have some symmetry principle to guide us in designing those new theories. Supersymmetry is such a principle. The basic idea is fairly simple to state. As the color SU(3) group of the standard model reshufﬂes the color states of a given quark ﬂavor among themselves and the weak SU(2) group mixes ﬁelds appearing in weak doublets (e.g., the left-handed electron and neutrino states), supersymmetry is a symmetry that involves changing bosonic and fermionic ﬁelds into one another. Supersymmetry solves the hierarchy problem in a most ingenious manner: Fermion and boson loop corrections to particle masses cancel one another exactly! Actually, almost all the ultraviolet (high-momentum) divergences of conventional quantum ﬁeld theory disappear in supersymmetric ﬁeld theories (i.e., theories invariant under supersymmetry transformations), a fact that we will show explicitly with a few examples later in the book. Unfortunately, exact invariance under supersymmetry implies that each existing particle should have a partner of the same mass and same quantum numbers but with a spin differing by one-half, which is obviously not observed in nature (there is no spin-zero or spin-one particle of the same mass as the electron, for example). However, there is no need to throw away the baby with the bath water. The same type of problem occurs with the standard model, which is based on symmetry under the direct group product SU (3)c × SU (2) L × U (1)Y (a brief review of the standard model will be provided in Chapter 15). We know that the world is not exactly invariant under this symmetry because this would imply that all gauge * †

At least at “low energy” relative to the Planck mass, approximately 1019 GeV. At energies comparable to the Planck mass, a theory of quantum gravity is presumably necessary, and then all bets are off. The Higgs particle is special in two respects. It is the only fundamental particle of spin zero predicted by the standard model, and it is also the only particle in the standard model that has not been observed yet.

CHAPTER 1

Introduction

3

bosons would be massless, in contradiction with the observed properties of the W ± and Z bosons. The solution in the standard model is provided by the spontaneous breaking of the symmetry via the Higgs mechanism. One might therefore hope to ﬁnd some similar trick to spontaneously break supersymmetry. It turns out that the situation in supersymmetry is a bit more challenging than in the standard model, as we will discuss in Chapter 14. In a different vein, something truly remarkable emerges from the simple idea of mixing bosons and fermions: Performing two successive supersymmetric transformations on a ﬁeld gives back the same ﬁeld but evaluated at a different spacetime coordinate than it was initially. Therefore, supersymmetry is intimately linked to spacetime transformations. This is unlike any of the internal symmetries of the standard model. In fact, if we impose invariance of a theory under local supersymmetric transformations, we ﬁnd ourselves forced to introduce new ﬁelds that automatically reproduce Einstein’s general relativity!* The resulting theory is called supergravity. Supersymmetry has been showing up in some unexpected places. A notable example is superstring theory. When building a string theory that contains both bosonic and fermionic ﬁelds, supersymmetry automatically appears. The fact that the theory exhibits invariance under supersymmetry is the reason it is called superstring theory. Supersymmetry also has been used extensively in many modern developments in mathematical and theoretical physics, such as the Seiberg-Witten duality, the AdS-CFT correspondence, the theory of branes, and so on. It is important to understand that, for now, supersymmetry is a purely mathematical concept. We do not know if it has a role to play in the description of nature, at least not yet. But it is very likely that we will ﬁnd out in the very near future. Indeed, the recent opening of the Large Hadron Collider at the Centre Europ´een de Recherche Nucl´eaire (CERN) near Geneva offers the exciting possibility of ﬁnally testing whether supersymmetry plays a role in the fundamental laws of physics. All this makes supersymmetry terribly exciting to learn. Unfortunately, it is quite difﬁcult to teach oneself supersymmetry despite the vast literature on the subject. This brings us to the purpose of this book.

1.2 What This Book Is and What It Is Not There are essentially three roadblocks to teaching oneself supersymmetry. The ﬁrst is the highly efﬁcient but at the same time horrendously confusing notation that permeates supersymmetry books and articles. It’s hard not to get discouraged by the *

In roughly the same way that making a U (1) global invariance local automatically leads to electromagnetism, for example.

4

Supersymmetry Demystified

dizzying profusion of indices that sometimes do not even seem be used consistently, μ as in χa , χ a , χ a˙ , χa˙ , (σ 2 )ab , (σ 2 )a˙ b˙ , σ¯ a a˙ , and so on. There is, of course, a very good reason for introducing all this notation: It helps immensely in constructing invariant quantities and making expressions as compact as possible. Unfortunately, this also makes it difﬁcult for newcomers to the ﬁeld because more effort must be put into making sense of the notation than into learning supersymmetry itself! It is possible to keep the notation very simple by not using those strange dotted indices and by not making any distinction between upper and lower indices, but the price to pay is that expressions are more lengthy, and some calculations become quite awkward. In itself, this is not such a big deal. But the goal of this book is not only to help you learn the rudiments of supersymmetry; it is also to provide you with the background required to “graduate” to more advanced references on the subject. And this necessitates familiarizing you with the notation of dotted and undotted spinors. For this reason, a compromise has been made. At ﬁrst, the simpler and less compact notation will be used to ensure that the fancy indices do not get in the way of learning the basic concepts. After a few chapters of practice, however, the more advanced notation will be introduced with an emphasis on the rationale behind it. A second difﬁculty in learning supersymmetry stems from the language used to describe the fermionic ﬁelds of the theory. Supersymmetry requires the use of Weyl or Majorana spinors, whereas most of us became familiar only with Dirac spinors in our introductory quantum ﬁeld theory classes. The next two chapters therefore will be entirely devoted to introducing these types of spinors and explaining their relation to Dirac spinors. A third difﬁculty is that there are actually two methods for constructing supersymmetric theories. As you might expect, one method, the so-called superspace or superﬁeld formalism, is more powerful and compact but also much more abstract and difﬁcult for a beginner than the second, down- and dirty-method. Alas, most references on supersymmetry use right away or very early on the superﬁeld approach. In this book we will start with the less efﬁcient but easier to learn approach and will only later introduce superﬁelds and superspace. Supersymmetry is a huge and formidably complex topic. In writing a (ﬁnite) book on the subject, one is faced with a gut-wrenching dilemma. Should one try to touch on as many aspects of the theory as possible, albeit in a necessarily superﬁcial manner? Or should one restrict oneself to a very small (almost inﬁnitesimal) subset of the theory with the goal of making the presentation, and especially the mathematical derivations, as thorough and detailed as possible? Well, for the present book, the decision was not difﬁcult because the goal of the Demystiﬁed series is to provide readers with self-teaching guides. This implies that the accent must be put on covering in depth the more basic concepts, even if it means paying the price of sacriﬁcing several advanced topics. For example,

CHAPTER 1

Introduction

5

we will not cover so-called extended supersymmetries (the meaning of which will become more clear later in this book), supergraphs, or applications in mathematical physics and superstring theory such as the Seiberg-Witten theory, BPS states, the AdS/CFT correspondence, and so on. Even the basic applications we will discuss cannot be given full justice in such a short book. For example, we will only be able to afford a cursory look at loop calculations in supersymmetry and to sample only some aspects of supersymmetry breaking and of the so-called minimal supersymmetric extension of the standard model (known affectionately as the MSSM). The purpose of the present work is therefore not to make anyone an expert on supersymmetry or even on any particular application of supersymmetry. It is rather meant as a gentle introduction to the basic concepts, as a springboard toward more advanced references, whether they are on string theory, mathematical physics, or phenomenological applications. The goal is to give you a solid understanding of the basic concepts so that you will be able to read and understand more advanced references (some of which will be suggested in Section 1.4). This brings us to the important question of prerequisites. It will be assumed that you have some basic knowledge of quantum ﬁeld theory, at the level of doing calculations of simple scattering amplitudes. If you need your memory to be refreshed, the book by McMahon30 in the Demystiﬁed series offers a gentle introduction to quantum ﬁeld theory, whereas Maggiore’s presentation28 is only slightly more advanced but very pedagogical. In my humble opinion, the books by Srednicki44 and Hatﬁeld24 are among the best references available on quantum ﬁeld theory. The book by Peskin and Schroeder37 is an invaluable complement. As always, the books by Weinberg48 are ﬁlled with deep and unique insights but are at a more advanced level than the previous references. Even though the essential elements of the standard model and the key formula for the calculations of amplitudes will be reviewed, they will be presented with the intention of refreshing your memory, not of actually teaching you the material. A very pedagogical introduction to the standard model is offered by Grifﬁths.21 At a slightly more advanced level, the two volumes by Aitchison and Hey2 are extremely well written and pedagogical. The many books by Greiner19 and collaborators are ﬁlled with detailed calculations and clear explanations. The intended audience for this book is therefore physicists with some basic background in quantum ﬁeld theory and in particle physics who want to learn the nuts and bolts of supersymmetry but ﬁnd themselves quickly stuck when reading on the subject because of the unfamiliar notation and the lack of computational details provided. We are pretty convinced that if you work your way through this book, you will have acquired the necessary basic tools to make other books or references on supersymmetry much easier to understand, and you will be able to make some progress in the study of whatever aspect of supersymmetry tickles your curiosity the most.

Supersymmetry Demystified

6

1.3 Effective Field Theories, Naturalness, and the Higgs Mass Let’s get back to the issue of the naturalness problem brieﬂy mentioned in Section 1.1. The fact that the problem disappears completely in a supersymmetric theory is one of the most appealing aspect of supersymmetry (SUSY), so it deserves further elaboration. In the last 30 years, our understanding of renormalization in the context of particle physics has undergone a radical shift (or a paradigm shift to use the oftabused expression). To explain the difference between the “old” school of thought and the “modern” point of view, it’s convenient to consider for regulator a “hard” cutoff on the integration of momenta,

∞

dk →

0

dk

(1.1)

0

Of course, such a regulator breaks almost all the symmetries usually encountered in particle physics and therefore is not practical, but for the purpose of building physical intuition about renormalization, nothing beats a good old momentum cutoff. The old view of the cutoff was that it had no physical meaning, that it was a purely mathematical sleight of hand introduced to make the intermediate steps of the calculations mathematically well-deﬁned and that, by consequence, it had to be taken to inﬁnity at the end of the calculation. A renormalizable theory is, by deﬁnition, a theory in which any result for physical processes can be made cutoff-independent in the limit → ∞ by redeﬁning, i.e., renormalizing, the ﬁnite number of parameters appearing in the lagrangian (the so-called bare parameters). By contrast, the modern point of view is that all the quantum ﬁeld theories we know (possibly with the exception of M theory) are to be treated as effective ﬁeld theories, i.e., theories that are approximations of more fundamental quantum theories, and in that context, the cutoff must be viewed as a physically meaningful parameter whose value should not be taken to inﬁnity. Instead, the ﬁnite value of the cutoff corresponds to the energy scale at which the effects of the “new physics” beyond the effective ﬁeld theory become important. From that point of view, there is nothing wrong with nonrenormalizable theories. As long as the processes that are computed involve particles with a typical external (nonloop) momentum p much smaller than the scale of the new physics N P , the nonrenormalizable contributions appear in the form of an expansion of the form n

cn

p N P

n

CHAPTER 1

Introduction

7

where the coefﬁcients cn are of order one. As long as the momentum p is much smaller than the scale of new physics, one can neglect all but the ﬁrst few terms to a given precision. Of course, the desired accuracy also dictates the number of loops used in the calculation, so there are two expansions involved in any calculation. Obviously, as p approaches N P , the whole effective ﬁeld theory approach breaks down, and one needs to uncover the more fundamental theory (which might itself be an effective theory valid only up to some higher scale of new physics). In some sense, effective ﬁeld theories are the “quantum equivalents” of Taylor expansions in calculus. The expression 1 − x + x 2 /2 − x 3 /6 is a good approximation to exp(−x) as long as x (the analogue of p/ N P ) is much smaller than 1. If we use x = 2, we are outside of the radius of convergence of the expansion, and we must replace the “effective” description 1 − x + x 2 /2 . . . by the more fundamental expression exp(−x). However, things are not as obvious in the context of a quantum theory because even if the external momenta are small relative to the scale of new physics, the loop momenta have to be integrated all the way to that scale. So it might seem at ﬁrst sight that no matter how small the external momenta are, since the loops will be sensitive to the new physics, the very concept of an approximate theory does not make sense in the context of quantum physics. However, the diagrams with external momenta much smaller than N P and internal momenta on the order of the scale of new physics are highly off-shell and, because of the Heisenberg uncertainty principle, are “seen” by the external particles as local interactions with coefﬁcients suppressed by powers of the scale of new physics. A detailed and very pedagogical discussion of this point is presented in the paper by Peter Lepage,26 who has pioneered the use of effective ﬁeld theories in bound states and in lattice quantum chromodynamics (QCD). An excellent introduction to effective ﬁeld theories is the paper by Cliff Burgess.9 Let us get back to the problem of naturalness. The key point to retain from the preceding discussion is that we must think of the cutoff as representing the scale of new physics, not as a purely mathematical artifact that must be sent to inﬁnity at the end of the calculation. Consider, then, an effective ﬁeld theory that is to be regarded as an approximation of a more general theory, for energies much smaller than the scale of new physics. The question is whether it is natural to have dimensionful parameters in the effective theory, masses in particular, that are much smaller than N P . Of course, at tree level, there is never any naturalness problem; we may set the parameters of the effective theory to any value we wish. It’s the loops that may cause some trouble (as always!). There is a naturalness problem if the dimensionful parameters of the effective ﬁeld theory are driven by the loops up to the scale of new physics. In that

Supersymmetry Demystified

8

Figure 1.1 One-loop correction to a fermion propagator.

case, the only way to keep these parameters at their small values is to impose an extreme ﬁne-tuning of the bare parameters to cancel the loop contributions. To be more speciﬁc, consider the electron mass in quantum electrodynamics (QED), viewed as effective ﬁeld theory of the electroweak model. The scale of new physics is therefore of order N P ≈ 100 GeV, which is much larger than the electron mass, m e ≈ 0.511 GeV. The question is whether, if we introduce loop corrections, it is natural for the renormalized electron mass to remain so small. The one-loop correction to the electron mass, depicted in Figure 1.1, is roughly δm e α

d 4 k k/ + m e αm e N P ln k 2 k 2 − m 2e 4π me

(1.2)

Naive power counting would lead us to expect a linear divergence, δm e α N P , but the linear term in k in the numerator actually vanishes on integration, leaving us with a much milder logarithmic divergence. Even if we set N P 100 GeV, we ﬁnd that the correction to the bare mass is of order δm e /m e ≈ 19% relative to its tree-level value. This is a small correction, and therefore, a small electron mass is natural in the context of QED as an effective ﬁeld theory. Let’s look at the masses of gauge bosons. This time, the two one-loop diagrams of Figure 1.2 contribute if the gauge group is nonabelian, whereas only the ﬁrst diagram exists if the gauge group is abelian. Schematically, the two diagrams give a correction to the gauge boson mass of the form δm 2g

= αg

d 4k

k2 + m (k 2 − m 2 )2

Figure 1.2 One-loop corrections to a nonabelian gauge ﬁeld propagator.

(1.3)

CHAPTER 1

Introduction

9

where m is the mass of whatever is circulating in the loop, and αg is the usual eg2 /4π , where eg is the gauge coupling constant. Simple power counting predicts this time a quadratically divergent result, but again, this contribution actually vanishes on integration. If the gauge symmetry is unbroken, the correction to the gauge boson mass actually vanishes identically, δm 2g = 0. If the gauge symmetry is spontaneously broken, the correction is not zero, but the quadratic divergence is still absent, and the leading correction is only logarithmically divergent, N P 2 2 δm g αg m ln (1.4) m Once more, the correction is small even if the scale of new physics is much larger than the physical mass of the gauge boson. It is not fortuitous if the leading divergences in both our examples cancel out, leaving a milder logarithmic divergence. In both cases, a massless theory has more symmetry than the massive theory. The theory of a massless fermion has a chiral symmetry that ensures that the fermion will remain massless to all orders of perturbation theory. If we start with a massive fermion, the loop corrections no longer vanish, but they must go to zero in the massless limit. This is why the one-loop correction to the fermion mass could not be of the form δm e N P because this would not vanish as the limit m e goes to zero. It is therefore chiral symmetry that “protected” us from the linear divergence. In the gauge boson case, it is obviously the gauge symmetry that is restored when the gauge boson mass goes to zero. Therefore, all corrections to the gauge boson mass must vanish as the limit m g goes to zero. This is why we did not generate a quadratic divergence, which would not have satisﬁed this criterion. This deﬁnition of naturalness has been formalized by ’t Hooft45 : A theory is natural if, for all its parameters p which are small with respect to the fundamental scale , the limit p → 0 corresponds to an enhancement of the symmetry of the system.

Obviously, his corresponds to our N P . Let us now turn our attention to the mass of a scalar particle, e.g., the Higgs boson in the standard model. One one-loop correction arises from the fermion loop shown in Figure 1.3, and it corresponds to the integral δm 2h α 2

d 4k

k2 + m ≈ α2N P (k 2 − m 2 )2

(1.5)

This time, taking the mass of the scalar to zero does not lead to any enhancement of symmetry, and we obtain a quadratically divergent correction. If we take the mass of the Higgs to be of order 102 GeV and the scale of new physics to be of

10

Supersymmetry Demystified

Figure 1.3 One-loop correction to the propagator of a scalar. The dotted lines represent the scalar particle propagator, and the loop is a fermion loop.

order the usual grand uniﬁcation scale of about 1015 GeV, we see that the one-loop correction is roughly 24 orders of magnitude larger than the tree-level value of m 2s (taking α ≈ 10−2 ). This means that the bare squared mass of the Higgs would have to be ﬁne-tuned to 24 digits to keep the renormalized mass equal to its tree-level value! In other words, the mass of the Higgs is given by m 2H = C1 − C2 , where both C1 and C2 are of order 1028 GeV2 and have the same ﬁrst 24 digits! This is a bit as if you were to throw a needle up in the air and watch it fall on a marble ﬂoor exactly on its tip and stay upright in that position! You certainly would be ﬂabbergasted! It would not be impossible that it did land in exactly the right way to have it be in equilibrium and then be undisturbed by any air current after it landed, but it would seem so unlikely as to be preposterous. You certainly would look for some explanation for why this happened. Maybe this part of the ﬂoor is not marble, and the needle penetrated the surface? Maybe there is a magnet under the surface, and the needle is magnetized in such a way as to stay upright? We are in a similar situation when pondering the mass of the Higgs in the standard model. There has to be some way to obtain a reasonable mass of the Higgs without invoking a mind-boggling accidental cancellation between two huge numbers! So why not simply say that the mass of the Higgs could be huge, of the order of 1015 GeV? Because there are other constraints that this time put an upper bound on the Higgs mass. For example, unitarity of the S-matrix leads to the constraint m h < 780 GeV (see Ref. 8 for more details). This leaves us with only one possible way out: The scale of new physics must be much smaller than 1015 GeV. As a criterion for absence of ﬁne-tuning, it is natural to demand that a one-loop correction must be smaller than the tree-level √ value. Applying this to the Higgs mass in the standard model leads to N P ≤ m h / α ≈ 1 TeV. This makes the scale of new physics very low and accessible experimentally. What could happen at this new scale? It could be revealed that the Higgs boson is not an elementary particle but instead a bound state, as in technicolor theories. Or maybe there are extra dimensions that can affect physics at the TeV scale, or again, maybe particles are not pointlike but rather are excitations of strings.

CHAPTER 1

Introduction

11

But let us assume that the Higgs still would be present as an elementary point particle in some new, more fundamental, four-dimensional theory superseding the standard model in the TeV region. In that case, it would seem that the naturalness problem still would be present. The scalar masses now would be driven to the scale of new physics beyond this new theory, and we would be back to our starting point. Unless this new theory would possess some symmetry that would protect even scalar masses from large corrections. And this is where supersymmetry makes it triumphant entrance! In theories with unbroken supersymmetry, quadratic divergences are altogether absent, so there is no naturalness problem! This makes supersymmetry a very attractive candidate for the new physics lurking around 1 TeV.

1.4 Further Reading After you have ﬁnished this book, you will have the necessary background (and hopefully, the hunger) to consult more advanced references. Here are some suggestions. We should ﬁrst mention the ground-breaking papers that gave birth to supersymmetry. Supersymmetry was independently discovered in the west16,33,39 and in the former Soviet Union.17,46 This was followed by the seminal work of Wess and Zumino, who constructed explicit four-dimensional supersymmetric quantum ﬁeld theories.50 For more details, the ﬁrst chapter of Ref. 47 is highly recommended. If you are interested in the phenomenology of the MSSM, the best book to read after the present volume is the one by Aitchison.1 It is presented at the same level as the present volume and complements it nicely. The following books are slightly more advanced but should offer no difﬁculty after you have worked your way through Supersymmetry Demystiﬁed. Baer and Tata5 offer a large number of cross sections and decay-rate calculations. The book by Bin´etruy8 is quite exhaustive (with almost 400 references!) and discusses cosmologic implications as well as the MSSM in great depth. Weinberg’s book47 on supersymmetry is, as is all his work, a model of clarity and a great source of insights. It covers not only the MSSM but also supergravity, Seiberg-Witten, supergraphs, and several other topics. Bailin and Love6 focus on supersymmetry in the context of superstring theory. A book by Dine12 also covers supersymmetry and its applications to superstring theory, but is more recent and has a wider scope (it covers Seiberg-Witten, cosmology, supergravity, and many other topics). Although old and, unfortunately, hard to ﬁnd, the book by M¨uller-Kirsten and Wiedemann31 is a real gem because of the large number of derivations presented in great detail. A classic is the book by Wess and Bagger,49 which is a good resource

12

Supersymmetry Demystified

for learning about supergraphs and supergravity but is more advanced than the previous references. In a different category, Ref. 15 contains a large number of the most important early research papers on supersymmetry and is an invaluable resource. In the same vein, Ref. 25 contains all the earlier Physics Reports on supersymmetry and remains a very useful reference. In addition to books, a large number of excellent review papers or lecture notes on supersymmetry are available. For more details concerning the phenomenological aspects of the MSSM, some excellent resources are Refs. 10, 29, 32, 36, and 38. If you are interested in the Seiberg-Witten theory, Refs. 47 and 3 are good places to start. A couple of more formal reviews are Refs. 27 and 51. Reference 43 focuses on supersymmetry breaking. Reference 13 explains in great detail how to do calculations with Weyl spinors and provides a very large number of detailed calculations relevant to the MSSM. This is far from being an exhaustive list (Ref. 8 alone contains almost 400 references!). In particular, very few of the original research papers are cited in the References, and for that we apologize to the large number of researchers who have discovered and reﬁned supersymmetry over the years. You are invited to consult the bibliographies of the books and review papers mentioned here for a more complete list of references.

CHAPTER 2

A Crash Course on Weyl Spinors

As mentioned in the Introduction, supersymmetric theories deal with Weyl or Majorana spinors, not the more familiar Dirac spinors. The ﬁrst step to understand supersymmetry (SUSY) is therefore to become at ease with those two types of spinors. This chapter will focus on Weyl spinors and their connection with Dirac spinors. Along the way, we will work out many identities that will be used repeatedly throughout this book and will deﬁne some notation that will prove very handy in the building of supersymmetric theories. This won’t be glamorous stuff—only in Chapter 5 will we construct our ﬁrst supersymmetric theory—but it is crucial to become proﬁcient with the notation established in the ﬁrst few chapters before tackling SUSY. Indeed, most physicists working in SUSY (whom we like to refer to as “superphysicists”) will tell you that the brunt of the difﬁculty in learning SUSY is not understanding the theory itself but rather becoming at ease with Weyl and Majorana spinors and the daunting notation used to describe them. It is therefore imperative to ﬁrst spend some time

Supersymmetry Demystified

14

familiarizing ourselves with this notation; once this is achieved, calculations in SUSY will be a breeze (well, almost).

2.1 Brief Review of the Dirac Equation and of Some Matrix Properties Let’s start by reviewing a few basic relations and deﬁning our convention for the Dirac matrices. The Dirac equation is γ μ Pμ = m

(2.1)

or, using Feynman’s “slash” notation, P / = m

Here, Pμ represents the differential operator i∂μ . Throughout this book we use the so-called natural units, in which c = = 1, unless stated otherwise. It is also useful to recall the Dirac lagrangian: LDirac = (γ μ Pμ − m)

(2.2)

where the bar over the spinor denotes the Dirac adjoint, deﬁned as ≡ † γ 0 . Equation (2.2) is actually not a lagrangian but a lagrangian density (because it must be integrated over spacetime to yield the action), but we will follow the incorrect but widespread convention of dropping the adjective density most of the time. There are several different representations for the Dirac matrices γ 0 and γ used in the literature. We choose the representation γ = 0

0 1

1 0

γ =

0 σ

−σ

0

where, of course, 1 stands for the 2 × 2 identity matrix. These can be used to deﬁne γ μ = (γ 0 , γ ) The corresponding quantity with a covariant index is deﬁned as γμ = ημν γ μ = (γ 0 , −γ )

CHAPTER 2

A Crash Course on Weyl Spinors

15

where ημν is the ﬂat spacetime metric for which we use the form ημν = diag(1, −1, −1, −1) This is sometimes referred to as the mostly minus convention as opposed to the mostly plus convention (−1, 1, 1, 1). Usually, particle physicists adopt the mostly minus convention, whereas the mostly plus metric is preferred by general relativists. It is always important to check the convention used when consulting a new reference. For the sake of completeness, let us also give the matrix γ5 , which will be needed shortly. We choose the following representation: γ5 =

1 0

0 −1

(2.3)

This is often called the chiral or Weyl representation. Recall that the Pauli matrices are given by σ1 =

0

1

1

0

σ2 =

0

−i

i

0

σ3 =

1

0

0

−1

We will make extensive use of the Pauli matrices, so it would be useful to recall some of their properties (which all can be checked using the explicit matrix representations). First, they are hermitian, i.e., (σ i )† = σ i . However, they do not behave so nicely under the separate operations of complex conjugation and transposition: (σ 1 )∗ = σ 1

(σ 2 )∗ = −σ 2

(σ 3 )∗ = σ 3

(σ 1 )T = σ 1

(σ 2 )T = −σ 2

(σ 3 )T = σ 3

(2.4)

Obviously, σ ∗ = σ T

(2.5)

The fact that σ 2 is singled out will play a key role later on. The product of any two Pauli matrices is given by σ i σ j = 1δ i j + i i jk σ k

(2.6)

Supersymmetry Demystified

16

where δ i j is the Kronecker delta, and i jk is the totally antisymmetric Levi-Civita symbol. To be explicit, 123 = 231 = 312 = 1

321 = 213 = 132 = −1

and

with all the other components being zero. The Levi-Civita symbol makes it possible to write the components of a cross-product between vectors in a compact manner.

×B

may be written in either of For example, the kth component of the vector A the following two forms:

× B)

k = i jk Ai B j (A = ki j Ai B j

(2.7)

Note that we are using Einstein’s summation convention throughout this entire book: Any repeated index is implicitly summed over. However, we will require only Lorentz indices to satisfy the rule that they must appear in the upper and lower positions to be contracted. From Eq. (2.6) one can calculate the commutator of two Pauli matrices to be [σ i , σ j ] ≡ σ i σ j − σ j σ i = 2i i jk σ k

(2.8)

The anticommutator is also easily computed: {σ i , σ j } ≡ σ i σ j + σ j σ i = 2δ i j 1

(2.9)

From this anticommutator, we directly obtain σ i σ j = −σ j σ i

for

i= j

(2.10)

(σ i )2 = 1 We noted earlier that σ 2 stands apart from the other two matrices when we consider complex conjugation or transposition. Let us focus on σ 2 again, for which we have σ 2 σ 1 = −σ 1 σ 2

σ 2 σ 3 = −σ 3 σ 2

σ 2σ 2 = σ 2σ 2

CHAPTER 2

A Crash Course on Weyl Spinors

17

the last one being, of course, trivial. What is interesting is that we can combine these results with Eq. (2.4) to get σ 2 σ T = −σ σ 2 ∗

σ σ = −σ σ 2

2

(2.11) (2.12)

Using the fact that (σ 2 )2 = 1, we get for free σ 2 σ σ 2 = −σ T = −σ ∗ σ 2 σ T σ 2 = σ 2 σ ∗ σ 2 = −σ

(2.13)

where the second line is trivially obtained from the ﬁrst line by sandwiching the latter between two σ 2 matrices. These identities will come in handy later on.

whose components are numbers, not matrices. We will Consider now a vector A

· σ σ j . It will prove useful to rewrite soon encounter an expression of the form A this as

· σ σ j = Ai σ i σ j A = Ai σ j σ i + Ai [σ i , σ j ] = Ai σ j σ i + 2i i jk Ai σ k = Ai σ j σ i − 2i jik Ai σ k

· σ − 2i( A

× σ ) j =σj A

(2.14)

2.2 Weyl versus Dirac Spinors Let us now write the four-component Dirac spinor in terms of two-component spinors: η

= χ

(2.15)

In this book we will use capital Greek letters for four-component spinors and lowercase Greek letters for two-component spinors. There is no standard notation for the symbols used for the upper and lower two-component spinors; some authors switch the η and the χ or use different Greek letters altogether.

Supersymmetry Demystified

18

The two-component spinors appearing in Eq. (2.15) are called Weyl spinors. The reason it makes sense to decompose a Dirac spinor this way is that, as we will see in more detail in Section 2.4, η and χ transform independently under Lorentz transformations; i.e., they do not mix. The technical way to express this is to say that a Dirac spinor forms a reducible representation of the Lorentz group, whereas Weyl spinors form irreducible representations. This implies that from a purely mathematical point of view, Weyl fermions may be considered more “fundamental” than Dirac spinors. The reason why one has to put together two Weyl spinors to form a Dirac spinor will be discussed in Chapter 4. Note that setting either η or χ equal to zero in a Dirac spinor yields eigenstates of γ5 : η η =+ γ5 0 0 γ5

0 0 =− χ χ

The eigenvalue of γ5 is called the chirality of the spinor. By a slight abuse of language, we will say that the upper two-component spinor η has a chirality of +1, whereas χ has a chirality of −1 (this is an abuse of language because it is actually the four-component spinor with two components set to zero that has a deﬁnite chirality, not the two-component spinors themselves). A Weyl spinor with positive chirality is sometimes referred to as a right-chiral spinor, and a Weyl spinor with negative chirality is referred to as a left-chiral spinor. However, some references replace the adjectives right-chiral and left-chiral by right-handed and left-handed, which is quite unfortunate because, in general, knowing the chirality does not tell us the handedness of a particle! We will soon discuss how the handedness or, to be more technical, the helicity of a spinor is deﬁned, but the key point to make here is that the notions of chirality and helicity of a spinor coincide only when a particle is massless. For a massive particle, both the left-chiral and the right-chiral states are, in general, linear combinations of left-handed and right-handed states. We will be careful about consistently using left-chiral and right-chiral to describe the eigenstates of the chirality operator. As we just mentioned, these two terms are synonyms of left-handed and right-handed only in the massless case. Why has it become common to use left-handed and right-handed to describe the states of deﬁnite chirality, even for massive particles? We don’t know for sure, but if we had to bet, our guess would be that people got so used to working with massless neutrinos in the standard model and talking about left-handed neutrinos and right-handed antineutrinos that they kept using this nomenclature even when

CHAPTER 2

A Crash Course on Weyl Spinors

19

they started considering massive neutrinos, and this habit has made its way into SUSY, where Weyl spinors are ubiquitous. However, we would not bet a lot of money. But why not forget about chirality and just work with helicity? The answer is that, as already mentioned, it is the chirality that tells us how a spinor behaves under Lorentz transformations! When a spinor is massive, it is its chirality that speciﬁes what representation of the Lorentz group it belongs to, not its helicity. Obviously, it is the behavior under Lorentz transformations that matters when constructing theories, so we will care much more about chirality than about helicity. Indeed, we will only discuss helicity in a few paragraphs in the entire book, whereas chirality will be omnipresent from the beginning to the end. We will discuss in detail the Lorentz transformations of the chiral states in Section 2.4. We will therefore show explicit subscripts R and L on the spinors η and χ for the remainder of the chapter to indicate their chirality. Of course, if the spinors are massless, the labels also tell us the helicity. When we start constructing supersymmetric theories, we will be working exclusively with left-chiral spinors, so subscripts won’t be necessary. Be warned that when it comes to the different types of spinors, there is no standard notation. Most books or papers do not show explicitly the indices L and R, and there is no general consensus on what Greek letters to use for leftand right-chiral spinors. Some switch the χ and η symbols for left- and rightchiral spinors, whereas others use different Greek letters altogether. In addition, some equations and expressions will be different if other representations of the gamma matrices are used. On top of that, some authors place the right-chiral spinor in the lower half of the Dirac spinor rather than at the top. Always check the notation and representation used when consulting a new reference. Caveat emptor! It is convenient to introduce the operators PR and PL , which, when applied to a full Dirac spinor, project out the right-chiral and left-chiral Weyl spinors. In other words, PR = PR

ηR χL

ηR = 0

PL = PL

ηR χL

0 = χL

(2.16)

which imply γ5 PR = PR

γ5 PL = −PL

(2.17)

Supersymmetry Demystified

20

The explicit representations of these operators are obviously PR =

1

0

0

0

PL =

0

0

0

1

It is easy to verify that with the representation of γ5 given in Eq. (2.3), we may write these projection operators as PR =

1 + γ5 2

PL =

1 − γ5 2

(2.18)

where now the 1 obviously represents the 4 × 4 identity matrix as opposed to the 2 × 2 identity matrix that it has represented so far. Most references are even less discerning than we are here and simply use the numeral 1 to represent the identity matrix in any dimension. It is also easy to check that these two operators satisfy the following properties: (PL )2 = PL (PR )2 = PR PL PR = PR PL = 0 γ5 PR = PR γ5 = PR γ5 PL = PL γ5 = −PL The ﬁrst three identities are, of course, typical for projection operators. If we substitute Eq. (2.15) into the Dirac equation, we obtain two coupled equations for the Weyl spinors: (E 1 − σ · p ) η R = m χL (E 1 + σ · p ) χL = m η R

(2.19)

Let us pause to introduce some new notation that will prove to be very useful later on. Let us deﬁne σ μ to be the following set of four matrices: σ μ ≡ (1, σ )

(2.20)

σ¯ μ ≡ (1, −σ )

(2.21)

We also will need to introduce

CHAPTER 2

A Crash Course on Weyl Spinors

21

It’s important to emphasize that the bar used here has nothing whatsoever to do with the one used to denote Dirac conjugation of a four-component spinor ﬁeld,

= † γ 0 . A different symbol over σ μ —maybe a tilde—would have been more appropriate, but this is standard notation, and we are stuck with it. Equation (2.13) imply the following identities: σ 2 (σ μ )T σ 2 = σ¯ μ σ 2 (σ¯ μ )T σ 2 = σ μ

(2.22)

If we take the transpose of both equations (and use (σ 2 )T = −σ 2 ), we also get σ 2 σ μ σ 2 = σ¯ μT σ 2 σ¯ μ σ 2 = σ μT

(2.23)

Since (σ i )∗ = (σ i )T , these equations are still valid with the transpose replaced by a complex conjugation. The four identities in Eqs. (2.22) and (2.23) will be so useful to us that they have been included in Appendix A, which lists the most important identities used throughout this book. An identity that is easy to prove (see Exercise 2.1) is σ μ σ¯ ν + σ ν σ¯ μ = 2ημν 1

(2.24)

Actually, doing the proof makes it obvious that the following also holds true: σ¯ μ σ ν + σ¯ ν σ μ = 2ημν 1

(2.25)

EXERCISE 2.1 Prove Eq. (2.24). Hint: Consider separately the cases μ = ν = 0, μ = 0 and ν = i, ν = 0 and μ = i and ﬁnally μ = i and ν = j (where it is understood that a Latin index may take a value of 1, 2, or 3 only). Using the deﬁnitions (2.20) and (2.21), the two coupled equations (2.19) may be written as Pμ σ μ η R = m χL Pμ σ¯ μ χL = m η R

(2.26)

Supersymmetry Demystified

22

We also could have obtained Eq. (2.26) directly from the Dirac equation by noting that γ μ may be written as γμ =

0

σ¯ μ

σμ

0

(2.27)

We also can write the Dirac lagrangian in terms of the Weyl spinors by substituting Eqs. (2.15) and (2.27) in (2.2): LDirac = η†R σ μ i∂μ η R + χL† σ¯ μ i∂μ χL − m η†R χL − m χL† η R

(2.28)

Note one important point: Despite the fact that σ μ and σ¯ μ carry a Lorentz index, they are, of course, not themselves four-vectors (for the same reason that γ μ is not a four-vector). However, we can form quantities with well-deﬁned Lorentz properties by sandwiching them between Weyl spinors, as we will discuss in more detail below.

2.3 Helicity Equation (2.26) shows clearly that the Dirac equation mixes η R and χL . We cannot have one without the other. For example, if we set η R = 0, the ﬁrst equation yields 0 = mχL , which implies that χL must be zero as well, unless the mass is zero. If the mass is zero, the two equations for η R and χL decouple to give σ · p η = ηR | p | R σ · p χ = −χL | p | L

(2.29)

where we have set E = | p |, which is valid for a massless particle (recall that in natural units, c = 1). So, for a massless particle, the Dirac equation actually breaks up into two independent equations acting on two different two-component spinors. These two equations are known as the Weyl equations. As we will discuss more later, this implies that when the mass is zero, we should think of η R and χL as ﬁelds describing two different particles instead of two states of a single particle. In fact, one can even build theories in which only one of the

CHAPTER 2

A Crash Course on Weyl Spinors

23

two is present, whereas the second one is absent altogether (this is of course the case in the standard model which contains only left-handed massless neutrinos). If we multiply those equations by /2 (showing explicitly for the moment)

we get and use the fact that σ /2 is the spin operator S, S · pˆ η R = η R 2 S · pˆ χL = − χL 2

The operator S · pˆ is called the helicity operator, and the corresponding eigenvalue is the helicity of the state. This eigenvalue corresponds to the component of the spin along the direction of the motion of the particle. We see that in the massless limit, the Weyl spinors are eigenstates of the helicity operator. The fact that η R has an helicity of /2 when m = 0 tells us that its spin is aligned with its direction of motion. If spin were a classical angular momentum, this would correspond to a particle following the path of a right-handed helix, which is why we refer to the two upper components of a massless Dirac spinor as a right-handed Weyl spinor. Obviously, χL corresponds to a left-handed Weyl spinor. There is a simple physical argument to explain why the two helicity states do not decouple for a massive state. To visualize, consider a particle with a classical angular momentum vector. Let’s say that the particle moves to the right while tracing a right-handed helix around the positive x axis. If we boost ourselves (the observers) to a frame that is moving faster than the particle, we will now observe it moving to our left while tracing a left-handed helix. Therefore, boosting to a frame moving faster than the particle changes its helicity state from right-handed to lefthanded, and vice versa. This has the implication that both states must be included to describe a massive spinor. The argument clearly fails when the particle is massless because it is impossible to boost to a frame moving faster than the particle. In that case, the two helicity states can never get mixed under Lorentz transformations, and they can be thought of as different particles altogether (as opposed to two states of a single particle in the massive case). It’s important to keep in mind that the chiral spinors χL and η R only have welldeﬁned helicities in the massless limit. Let’s call ψ+ and ψ− the states of positive and negative helicities (i.e., the right-handed and left-handed states). Then, in general, we have χL = C1 ψ+ + C2 ψ− η R = C2∗ ψ+ − C1∗ ψ−

(2.30)

Supersymmetry Demystified

24

which can, of course, be inverted to give ψ+ =

C1∗ χL + C2 η R |C1 |2 + |C2 |2

ψ− =

C2∗ χL − C1 η R |C1 |2 + |C2 |2

(2.31)

In other words, chirality and helicity offer two different bases on which the spinors may be expanded. The massless limit corresponds to C1 being identically 0 and C2 = 1. Helicity, the projection of the spin on the direction of motion of the particle, is easier to picture physically than the abstract concept of chirality. But we care more about the chiral eigenstates χL and η R because, as mentioned earlier, these are the states that have well-deﬁned Lorentz transformation properties, as we will discuss next.

2.4 Lorentz Transformations and Invariants Our next step is to look at the Lorentz transformations of spinors. Knowing how quantities transform under rotation and boosts is obviously crucial to building invariant lagrangians. The inﬁnitesimal transformations of the right-chiral and leftchiral Weyl spinors are

i 1

η R → η R = 1 + · σ − β · σ η R 2 2 i 1 χL → χL = 1 + · σ + β · σ χL 2 2

(2.32)

where is the inﬁnitesimal rotation vector (the direction of gives the axis of rotation, and its magnitude gives the amount of rotation), and β is the inﬁnitesimal boost parameter. The transformations [2.32] show explicitly that, as we mentioned earlier, the left-chiral and right-chiral spinors transform independently under Lorentz transformations. They are therefore inequivalent irreducible representations of the Lorentz group and are the fundamental “building blocks” from which any other spinor representation can be constructed. Because the Lorentz algebra is equivalent to the algebra su(2) × su(2), all representations can be labeled by two numbers that are multiples of one-half. The two

CHAPTER 2

A Crash Course on Weyl Spinors

25

fundamental representations then are described by the quantum numbers ( 12 , 0) and (0, 12 ), which describe, respectively, the left-chiral and right-chiral spinors. We won’t use this language in this book, but it is important to be aware of this notation because many SUSY references use it. By the way, even though the algebra acting on the Weyl spinors is su(2) × su(2), the group under which they transform is not SU (2) × SU (2) but instead S L(2, C), the universal covering group of the Lorentz group S O(3, 1). This might be useful to know if you want to impress your friends. For the sake of comparison, consider the corresponding transformations for a four-vector, e.g., the four-momentum P μ . The zeroth component P 0 = E and the vector components p transform under rotations and under boosts in the following way: E → E = E − β · p

p → p = p − × p − β E

(2.33)

Now, let us ask: What Lorentz-invariant quantity can we build out of spinors? We can use the Lorentz transformations (2.32) to ﬁgure this out or simply use the Dirac lagrangian Eq. (2.2) as a guide. Let us ﬁrst focus on the mass term. The Dirac lagrangian tells us that combination † γ 0 is invariant. Writing this in terms of Weyl spinors, we ﬁnd that

= † γ 0 = η†R χL + χL† η R is invariant under Lorentz transformations. Actually, it turns out that each term on the right-hand side is separately invariant. EXERCISE 2.2 Show that η†R χL is invariant under Lorentz transformations. Note that the Lorentz transformations of the spinors are given to ﬁrst order in and β so that all terms of higher order may be dropped. Now consider the kinetic energy term of the Dirac lagrangian γ μ Pμ . Since this is invariant and Pμ is a four-vector, the quantity γ μ must transform as a fourvector (as a contravariant four-vector, to be precise). If we rewrite this expression in terms of Weyl spinors and use Eq. (2.27), we ﬁnd that

γ μ = η†R σ μ η R + χL† σ¯ μ χL transforms as a four-vector under Lorentz transformations. Actually, it turns out that each term on the right separately transforms as a four-vector. As noted earlier, although σ μ and σ¯ μ carry a Lorentz index, they are not themselves four-vectors.

Supersymmetry Demystified

26

The matrices σ μ must be sandwiched between an η†R and an η R to yield a four-vector, and σ¯ μ must be sandwiched between χL† and χL . EXERCISE 2.3 Prove that η†R σ μ η R transforms as a four-vector under a Lorentz transformation. Hint: Consider ﬁrst η†R σ 0 η R and show that it transforms like the energy E in Eq. (2.33). Then show that η†R σ i η R transforms like the momentum three-vector.

2.5 A First Notational Hurdle Here we should pause to discuss a ﬁrst (of many!) possible source of confusion owing to notation. This ﬁrst hurdle concerns the use of the hermitian conjugate (or “dagger”) symbol, †. In linear algebra, taking the hermitian conjugate means taking the complex conjugate and transpose of a matrix or vector. However, in quantum ﬁeld theory, the symbol † has in addition a second role: When applied to a quantum ﬁeld, it consists of taking a complex conjugate of any complex number as well as replacing annihilation operators by creation operators (and vice versa). For example, writing a free complex scalar ﬁeld φ(x) as φ(x) =

d 3k [a(k)e−ik·x + b† (k)eik·x ] 2E(2π )3

the hermitian conjugate of the ﬁeld is given by φ † (x) =

d 3k [a † (k)eik·x + b(k)e−ik·x ] 2E(2π )3

(2.34)

Clearly, if the dagger is applied to a scalar ﬁeld, it only represents taking the hermitian conjugate in the “quantum ﬁeld operator sense” shown in Eq. (2.34). Consider now quantum ﬁelds that are themselves components of column or row vectors, e.g., the spinor χ . Both components of χ —let’s call them χ1 and χ2 — are quantum ﬁelds, and it makes sense to talk about χ1† and χ2† . This would be nonsensical for components which are simple complex numbers because we can’t take the hermitian conjugate of a complex number. In the case of quantum spinor ﬁelds, taking the dagger of a component has a clear meaning: It stands for the hermitian conjugation in the quantum ﬁeld operator sense (complex conjugation plus exchange of creation and annihilation operators). The problem arises when we consider something like χ † . What is meant by this is not completely clear, a priori. Does this represent the column vector with the

CHAPTER 2

A Crash Course on Weyl Spinors

27

hermitian conjugation in the quantum ﬁeld operator sense applied to the components, namely, † χ1 χ† = (2.35) χ2† or is the hermitian conjugate applied in both the operator and the linear algebra sense, as in χ † = χ1† , χ2† Most authors simply use χ † to represent both expressions, letting the context dictate which meaning is implied. With some practice, it is not difﬁcult to get used to this double role of the dagger, especially when expressions are fairly short, but it is distracting to have to worry about making sense of an ambiguous notation when learning a new subject. For this reason, we will use the following convention in this book: When a dagger will be applied to a spinor ﬁeld, it will represent the hermitian conjugation of each component of the spinor ﬁeld in the quantum ﬁeld operator sense as well as a transpose. Therefore, the quantity χ † will always represent the † row vector with components χi , i.e., χ † = χ1† , χ2† Note that this is the convention we used in the preceding section. The question is, then, what do we do if we want to represent the column vector (2.35) unambiguously? We will simply have to “undo” the transpose included in the linear algebra dagger, so we will write χ

†T

≡

χ1† χ2†

(2.36)

Of course, this notation would never be used if we were dealing with vectors made of complex numbers because we would simply write χ ∗ instead of χ †T . Here, the quantum operator dagger is applied to each component of the spinor, which is why we cannot think of χ †T as χ ∗ . This notation is admittedly a bit awkward, but it is a small price to pay for the gain in clarity we get by explicitly distinguishing the column and row spinors χ †T and χ † . Note that the possible confusion arises only when the dagger is applied to spinor quantum ﬁelds. When applied to matrices such as Dirac matrices or Pauli matrices,

28

Supersymmetry Demystified

the dagger obviously stands for the usual hermitian conjugation of linear algebra, and when applied to a scalar ﬁeld, it clearly represents the hermitian conjugation in the quantum ﬁeld operator sense.

2.6 Building More Lorentz Invariants Out of Weyl Spinors The simplest supersymmetric theory, which we will construct in Chapter 5, involves a single Weyl spinor (chosen by convention to be left-chiral) coupled to scalar ﬁelds. Later we will work with theories containing several left-chiral spinors. It is therefore important to ﬁgure out all the possible ways to construct Lorentz-invariant terms out of left-chiral spinors only. For the sake of completeness, we also will write down Lorentz-invariant expressions containing only right-chiral spinors or mixing leftand right-chiral spinors, although these won’t be needed in the rest of this book. Consider, then, a single left-chiral spinor χL . We already know one Lorentzinvariant expression that contains only χL : the kinetic term appearing in Eq. (2.28). On the other hand, the mass term that we obtained from the Dirac lagrangian mixes a left- and right-chiral spinor. So an interesting question arises: Is it possible to build a mass term out of a left-chiral spinor only? It turns out that it is possible, but not completely trivial. We know that χL† η R is invariant, so what we need is to somehow build out of the left-chiral ﬁeld χL something that transforms like a right-chiral ﬁeld η R ! If we then multiply χL† by this new quantity, we will have obtained a Lorentz-invariant mass term for the left-chiral ﬁeld. As a ﬁrst guess to what this quantity could be, let’s consider χL†T . Recall from the preceding section that this is a column vector whose components are the dagger of the quantum ﬁelds [see Eq. (2.36)]. Keeping in mind that the hermitian conjugate is applied only to the quantum ﬁelds and that only a complex conjugation is applied to everything else, and using Eq. (2.32), we ﬁnd that χL†T transforms as ∗ i 1 χL†T → (χL†T ) = (χ L )†T = 1 + · σ + β · σ

χL†T 2 2 i 1 ∗ ∗

= 1 − · σ + β · σ

χL†T 2 2 Unfortunately, this does not transform like a right-chiral spinor [see ﬁrst line of Eq. (2.32)]. There are two problems: The two signs are wrong, and the Pauli matrices are complexed conjugated. But it turns out that both problems can be solved in one

CHAPTER 2

A Crash Course on Weyl Spinors

29

shot if we recall Eq. (2.12): σ 2 σ ∗ = −σ σ 2 which is exactly what the doctor ordered! To see why this helps, consider the Lorentz transformation of the quantity iσ 2 χ L (the inclusion of a factor of i is purely conventional but convenient when we relate particle and antiparticle states, as we will see in Chapter 4): †T

iσ χ L 2

∗ 1 i 1 + · σ + β · σ χ †T L 2 2 1 i 2 ∗ ∗

= iσ 1 − · σ + β · σ χ †T L 2 2 i 1 = 1 + · σ − β · σ iσ 2 χ †T L 2 2

→ iσ 2 χ †T = (iσ 2 )∗ L

which is the transformation of a right-chiral spinor! We have therefore built something that transforms like a right-chiral spinor out of a left-chiral spinor. This is a key result, so let us emphasize it: iσ 2 χ †T transforms like a right-chiral spinor L

(2.37)

The matrix iσ 2 will appear in so many of our calculations that it is useful to summarize some of the properties that will often be used: (iσ 2 )T = −iσ 2

(iσ 2 )∗ = iσ 2

(iσ 2 )† = −iσ 2

(iσ 2 )2 = −1

Now we are ready to build Lorentz invariants that contain only left-chiral spinors. We consider only invariants with no Lorentz indices for now. Since the expression η†R χL is invariant under Lorentz transformations and iσ 2 χ †T transforms like η R , we L know that the following expression is also invariant: 2 †T † iσ χL χL = χLT (iσ 2 )† χL = χLT (−iσ 2 )χL

(2.38)

This is what we were seeking: a Lorentz invariant mass term containing only leftchiral spinors!

Supersymmetry Demystified

30

Since we know that χL† η R is invariant, we obtain a second invariant containing only left-chiral spinors: χ †L iσ 2 χ †T L

(2.39)

The two invariant combinations (2.38) and (2.39) play such an important role in the construction of supersymmetric theories that they are assigned special notations that we will discuss shortly. Before doing that, let us build other invariants made of right-chiral spinors alone or mixing left- and right-chiral spinors. First, we need to build something out of the right-chiral spinor η R that will transform like a left-chiral spinor. It is easy to check that −iσ 2 η†T transforms like a left-chiral spinor R

(2.40)

Again, the reason why we choose to include a factor of −i will become clear when we discuss antiparticles in Chapter 4. To build other invariants, we could proceed again by relying on the invariance of the terms appearing in the Dirac lagrangian. However it is also instructive to write down explicitly the Lorentz transformation properties of the various quantities we are dealing with. Let’s deﬁne the matrix A as A ≡1+i

β · σ

· σ

− 2 2

With this deﬁnition, right-chiral spinors have for Lorentz transformation η R → A ηR . Using the fact that the Pauli matrices are hermitian, we can easily ﬁnd the hermitian conjugate of the matrix A: A† = 1 − i

· σ

β · σ

− 2 2

The inverse matrix is obtained simply by changing the sign of and η because these are inﬁnitesimal quantities. Thus we ﬁnd A−1 = 1 − i

· σ

β · σ

+ 2 2

CHAPTER 2

A Crash Course on Weyl Spinors

31

In turn, this implies (A−1 )† = (A† )−1 = 1 + i

· σ

β · σ

+ 2 2

where we stressed that the order of the dagger and inverse operations does not matter. This is the matrix that appears in the transformation of left-chiral spinors. Thus the quantities we have at our disposal, together with their behavior under Lorentz transformations, are η R → Aη R †T †T iσ 2 χ L → A iσ 2 χ L χL → (A† )−1 χL −iσ 2 η†T → (A† )−1 − iσ 2 η†T R R

(2.41)

It is then clear how to build invariants. If we call type I the expressions that transform with the matrix A and type II those that transform with the matrix (A† )−1 , then the possible invariants are either of the form (type I)† (type II) or of the form (type II)† (type I) This gives eight combinations (counting as different combinations related by hermitian conjugation), but it turns out that only six of those are independent, which may be taken to be − χLT iσ 2 χL χL† η R − η†R iσ 2 η†T R

†T

χL† iσ 2 χ L η†R χL η TR iσ 2 η R

(2.42)

The ﬁrst two expressions involve only the left-chiral spinors that we obtained earlier. The next two expressions mix the two types of spinors and are the ones appearing in a Dirac mass term. The last two expressions contain only right-chiral spinors.

Supersymmetry Demystified

32

EXERCISE 2.4 Show that the two remaining combinations, involving both iσ 2 χL†T and −iσ 2 η†T , R are equivalent to two of the expressions already listed.

2.7 Invariants Containing Lorentz Indices Let us now consider terms that contain derivatives of spinor ﬁelds. We have seen, by looking at the Dirac equation written in terms of Weyl spinors [Eq. (2.26)], that (after using Pμ = i∂μ ) σ μ i∂μ η R transforms like a left-chiral spinor

(2.43)

σ¯ μ i∂μ χL transforms like a right-chiral spinor

(2.44)

whereas

Note how the matrix σ μ “belongs” with right-chiral spinors and σ¯ μ “belongs” with left-chiral spinors, in the sense that these are the combinations required to build quantities that have well-deﬁned Lorentz transformation properties. Applying a σ μ to a left-chiral spinor does not give anything with well-deﬁned Lorentz properties, so it’s not a useful thing to do. However, applying a σ¯ μ to a left-chiral spinor and then applying a σ μ does produce something interesting, as will be illustrated in Exercise 2.5. To build Lorentz invariants containing derivatives, we simply have to use the expressions listed in Eq. (2.42) with the χL appearing in those equations replaced by Eq. (2.43) or the η R replaced by Eq. (2.44). As examples, consider building invariants containing only left-chiral spinors and only one derivative. These are obtained by taking the terms in Eq. (2.42), which contain both χL and η R , and replacing η R by Eq. (2.44). If we do this for the term χL† η R of Eq. (2.42), we get χL† σ¯ μ i∂μ χL

(2.45)

which we recognize as the kinetic term of the left-chiral spinor that appears in the Dirac lagrangian [Eq. (2.28)]. A second possibility is the term η†R χL , which, after replacing η R by Eq. (2.44), gives

† σ¯ μ i∂μ χL χL = −i ∂μ χL† σ¯ μ χL

(2.46)

CHAPTER 2

A Crash Course on Weyl Spinors

33

where we have used the fact that (σ¯ μ )† = σ¯ μ because the Pauli matrices are hermitian. As part of a lagrangian density, however, an integration by parts can be used to show that the term in Eq. (2.46) is equivalent to Eq. (2.45) after discarding a surface term. We will frequently use integrations by parts on expressions appearing in lagrangian (lagrangian densities, to be exact). We will always assume that the surface terms vanish so that we may write

d x A ∂μ B = − 4

d 4 x (∂μ A)B

where A and B are arbitrary expressions containing quantum ﬁelds. This then shows clearly that the right-hand side of Eq. (2.46) is equal to Eq. (2.45). We also can construct invariants containing two derivatives of a left-chiral ﬁeld by replacing both η R in the last two terms of Eq. (2.42) by Eq. (2.44). We will never have to deal with these terms, however, because they are not renormalizable (an issue that we will address in Chapter 8). But there are plenty of other things of interest that we can build with spinors: Lorentz scalars, vectors, and tensors! Exercise 2.5 offers two examples. EXERCISE 2.5 Build a contravariant vector (i.e., with one upper Lorentz index) and a second rank contravariant tensor using two left-chiral spinors and no derivatives. In principle, this is all we need to start studying SUSY! Unfortunately, in practice, we need to introduce much more notation. However, as mentioned in the Introduction, the choice has been made to keep the notation as simple as possible until Chapter 11 in order to keep it from getting in the way of understanding the basic concepts of SUSY. Nevertheless, we will cheat a little and introduce a tiny bit of extra notation that will make many equations far less cumbersome. Before doing this, however, let’s pause to mention a simple identity that will help us countless times.

2.8 A Useful Identity One of the identities that will be used the most often is a simple modiﬁcation of a relation familiar from elementary linear algebra. Consider two plain column vectors v and w whose components are ordinary complex numbers (“ordinary” in the sense that they are commuting quantities) and a matrix A whose elements are also complex numbers. Then the following equality

34

Supersymmetry Demystified

holds: v T Aw = w T A T v

(2.47)

This can be checked easily by writing the indices explicitly. This identity follows from the fact that v T Aw = (v T Aw)T which is true because the quantity v T Aw is just a number, so it is trivially equal to its own transpose. Applying the the transpose to the parenthesis on the right-hand side, we recover Eq. (2.47). Of course, if complex conjugates are taken on any of the quantities, they will appear on both sides of the equation. For example, v † A∗ w ∗ = w † A† v ∗ We will not be dealing with vectors with complex numbers as components but with spinors, whose components are Grassmann quantities, which anticommute. In that case, we must introduce a minus sign when we switch the order of the spinors so that Eq. (2.47) is replaced by α T A β = −β T A T α

(2.48)

where α and β are arbitrary spinors. Of course, if a hermitian conjugate is applied to either spinor or to the matrix, we need to take this into consideration. For example, α T A β †T = −β † A T α We will use Eq. (2.48) or its equivalent form with hermitian conjugates applied to some of the terms repeatedly throughout this book.

2.9 Introducing a New Notation We have achieved our goal for this chapter, which was to build Lorentz invariants out of Weyl spinors. In principle, before moving on, we should discuss the shorthand notation of dotted and undotted indices, also called the van der Waerden notation, used by “superphysicists” when they do calculations in SUSY. This notation is very useful

CHAPTER 2

A Crash Course on Weyl Spinors

35

in that it makes it much easier to identify Lorentz invariant combinations of Weyl spinors, without the need to remember where to put the iσ 2 matrices or where to place hermitian conjugates. Unfortunately, it is also a real pain for beginners because each spinor now comes in four different versions. Of course, we could avoid altogether this shorthand notation because it is not absolutely necessary to do SUSY calculations. But there would be two major drawbacks to this. One is that it would be more difﬁcult for you to consult other publications on SUSY if you haven’t been exposed to this notation, which is used widely. The second problem is that it does make calculations much simpler when we have to deal with complicated expressions (and expressions do get fairly messy in SUSY!). However, for the ﬁrst several chapters, our calculations will be simple enough that using the shorthand notation would not provide a noticeable advantage and would make the topic harder to learn. On the other hand, starting in Chapter 11, where we begin working in superspace, calculations would become awkward if we would not take advantage of the shorthand notation of dotted and undotted indices. So we have opted for a compromise. The van der Waerden notation is introduced in Chapter 3 because it is a logical extension of the material covered in Chapter 2, but it will not be used until Chapter 11 (aside from part of an exercise in Chapter 6). By the time we reach Chapter 11, the basic concepts and tools of SUSY hopefully will be sufﬁciently familiar that it should not be too much of a problem to add a new layer of notation. We would therefore invite the reader to skip Chapter 3 on a ﬁrst reading and to get back to it before tackling superspace, in Chapter 11. In the rest of this chapter we will only be introduced to the bare minimum of extra notation needed until Chapter 11. In practice, SUSY theorists work almost exclusively with left-chiral spinors, even when considering supersymmetric extensions of the standard model (using some tricks that will be explained later), so we really only need to focus on the two invariants built out of a left-chiral spinor: χL† iσ 2 χL†T

χLT (−iσ 2 )χL

(2.49)

From now on we will drop the subscript L; it will be implicit that χ always stands for a left-chiral spinor. We deﬁne two new types of spinor dot products between left-chiral Weyl spinors as a shorthand notation for the two Lorentz invariants of Eq. (2.49): χ · χ ≡ χ T (−iσ 2 )χ χ¯ · χ¯ ≡ χ † iσ 2 χ †T

(2.50)

Supersymmetry Demystified

36

Let us write the two types of dot products explicitly in terms of the components of the spinors. It is conventional to write the components of a left-chiral Weyl spinor with lower indices, namely,

χ1 χ ≡ χ2

in terms of which the ﬁrst invariant of Eq. (2.50) is χ · χ = χ2χ1 − χ1χ2 At ﬁrst sight, this may seem to be identically zero, but recall that the χi are fermionic quantum ﬁelds, which anticommute. This is clear if we write the ﬁelds in terms of creation and annihilation operators. But what if we use the path integral approach, where quantum ﬁelds are not represented by operators? In that case, as you surely remember, the fermionic degrees of freedom must be represented by Grassmann variables, which also anticommute. In either case, we have χ1 χ2 = −χ2 χ1 ,

(χ1 )2 = (χ2 )2 = 0

Using the ﬁrst of these two relations, we get three equivalent expressions for the dot product χ · χ : χ · χ = χ2 χ1 − χ1 χ2 = 2 χ 2 χ1 = −2 χ1 χ2

(2.51)

Consider now the second invariant of Eq. (2.50). It is simply given by χ¯ · χ¯ = χ †1 χ †2 − χ †2 χ †1 = 2 χ †1 χ †2 = −2 χ †2 χ †1

(2.52)

This is it. This is all we are going to introduce in terms of new notation in this section. In the next chapter, a different way to represent these dot products will be introduced, but for now, you should think of χ · χ and χ¯ · χ¯ purely as shorthand notations for the expressions appearing on the right of Eq. (2.50), nothing more.

CHAPTER 2

A Crash Course on Weyl Spinors

37

To be honest, SUSY afﬁcionados usually don’t write any dot symbol (·) between the spinors, but it makes expressions more clear to include it, so that’s what we will do for the rest of this book. However, even the notation used here may cause some confusion, so let us issue a few warnings. First, note that the bar appearing over the Weyl spinors has nothing whatsoever to do with the bar used to represent Dirac conjugation for a four-component spinor,

= † γ 0 . We will discuss the bar notation for Weyl spinors in more depth later. For now, consider the bars in the second line of Eq. (2.50) to be simply a notation to distinguish the two types of dot products. Second, those “spinor dot products” have nothing to do with the usual dot products between four-vectors or three-vectors familiar from introductory physics. It may be tempting to think of χ · χ , say, as χ12 + χ22 , but this, of course, would be incorrect (and equal to zero!). If we want to calculate these dot products explicitly in terms of components, we need to go back to their explicit deﬁnitions given in Eq. (2.50). Third, it’s very important to keep in mind that the difference between the two types of dot products is not only that one involves the hermitian conjugate of the spinors, whereas the other does not; there is also a sign difference between the two deﬁnitions, as is shown explicitly in Eqs. (2.51) and (2.52). However, the two types of dot products obviously can be written in terms of one another using χ¯ · χ¯ = −χ †T · χ †T

(2.53)

The right-hand side is admittedly rather ugly, which is why the bar notation of the left-hand side is preferred. The point of showing this relation is that the bar notation is redundant; we could write everything in terms of unbarred spinors, at the price of sometimes getting awkward expressions. From Eq. (2.51), we see that the product of the two components of a spinor is related to the dot product of that spinor with itself in a very simple way: 1 χ1 χ2 = −χ2 χ1 = − χ · χ 2 whereas, of course, χ12 = χ22 = 0. These four relations can be summarized into a very useful identity: 1 χa χb = − (iσ 2 )ab χ · χ 2

(2.54)

Supersymmetry Demystified

38 Similarly, we have

1 χ †1 χ †2 = −χ †2 χ †1 = χ¯ · χ¯ 2 whereas (χ †1 )2 = (χ †2 )2 = 0, so the equivalent of Eq. (2.54) for the hermitian conjugated components is 1 † χa† χb = (iσ 2 )ab χ¯ · χ¯ 2

(2.55)

There is something important to repeat here: The hermitian conjugate in the last two equations is applied to the components of the spinor, not to the spinor itself. In that context, the dagger necessarily means the hermitian conjugate in the quantum ﬁeld operator sense. If the components were simple complex numbers, it would make no sense to write, say, χ1† . We would simply write χ1∗ . Keep in mind that † whenever we write components of a spinor with a dagger applied, as in χa , the quantum ﬁeld operator sense is always implied. There is another simple relation between the two types of products [much more elegant than Eq. (2.53)], but before demonstrating it, we ﬁrst have to deal with a tricky issue. Consider taking the hermitian conjugate of χ · χ : (χ · χ)† = (2χ2 χ1 )† We face a subtle problem. We know that the hermitian conjugate of the product of two matrices exchanges their order, but what do we do when we have Grassmann quantum ﬁelds (in other words, spinor quantum ﬁelds)? Do we include an extra minus sign owing to the exchange? The convention is to deﬁne the hermitian conjugate to exchange their order without introducing a factor of minus one. This is so important that we cannot resist the urge to put it in a box: The hermitian conjugate of the product of two spinor ﬁelds exchanges their order without introducing a minus sign

(2.56)

Again, an ambiguity arises if we use the path integral approach and write the fermion ﬁelds as Grassmann variables. In that case, instead of the hermitian conjugate of a product of ﬁelds, we would simply take the complex conjugate of that product. But, in order to be consistent with Eq. (2.56), we must deﬁne the complex

CHAPTER 2

A Crash Course on Weyl Spinors

39

conjugate of a product of Grassmann variables to change their order without introducing a minus sign! In other words, we deﬁne (αβ)∗ = β ∗ α ∗

(2.57)

where α and β are some arbitrary Grassmann variables. With this convention, we get (χ · χ )† = 2(χ2 χ1 )† = 2χ1† χ2†

(2.58)

which is precisely χ¯ · χ! ¯ So we have the following two important identities: (χ · χ )† = χ¯ · χ¯ † χ¯ · χ¯ = χ · χ

(2.59)

where the second line is simply the hermitian conjugate of the ﬁrst line. So far we have only written invariants built out of a single left-chiral spinor. Of course, we also will need to consider invariants made of two different left-chiral spinors, let’s say χ and λ. In this case, the two possible invariants mixing the two spinors are obviously χ ·λ

χ¯ · λ¯

At ﬁrst sight it seems that we also should consider λ · χ and λ¯ · χ, ¯ but it turns out that the order of the spinors in spinor dot products does not matter, i.e., λ·χ =χ ·λ λ¯ · χ¯ = χ¯ · λ¯

(2.60)

We have emphasized these identities by placing them in a box because they will be used repeatedly. They probably look wrong to you because the components of spinors anticommute, so shouldn’t we add a minus sign when we switch their order? The answer is no because the dot products used here involve the matrix σ 2 , which is antisymmetric. This introduces an extra minus sign that cancels the minus sign arising from the exchange of the spinor components. The following exercise will make this more explicit.

40

Supersymmetry Demystified

EXERCISE 2.6 Prove Eq. (2.60). Hint: Use the deﬁnitions (2.50) and the identity (2.48). The generalization of Eq. (2.59) to the products between two different spinors is easily found to be (λ · χ )† = λ¯ · χ¯ (λ¯ · χ¯ )† = λ · χ

(2.61)

Consider now three left-chiral spinors χ , α, and β. The following identities are easy to prove: 1 χ · αχ · β = − χ · χ α · β 2 1 χ¯ · α¯ χ¯ · β¯ = − χ¯ · χ¯ α¯ · β¯ 2

(2.62)

We see that the net effect of these identities is to swap the spinors appearing in the dot products. Such identities are generically referred to as Fierz identities. These particular examples of Fierz identities will prove to be very useful later in this book. EXERCISE 2.7 Prove Eq. (2.62). Hint: You may simply expand out completely both sides in terms of the components of the spinors and then move some of them around, or you may use Eqs. (2.54) and (2.55). In the next chapter we will introduce the shorthand notation of dotted and undotted indices mentioned in the Introduction, but as also mentioned there, we will not make use of it until Chapter 11. So feel free to go directly to Chapter 4 and come back to Chapter 3 later.

2.10 Quiz 1. Write χ¯ · λ¯ explicitly in terms of the components of the spinors. 2. Under what condition does a Weyl spinor have a deﬁnite helicity? Under what condition does it have a deﬁnite chirality?

CHAPTER 2

A Crash Course on Weyl Spinors

41

3. Do left- and right-chiral spinors transform the same way or differently under rotations? What about their transformations under boosts? 4. Write down the expression for a Dirac mass in terms of Weyl spinors. 5. Consider the two components χ1 and χ2 of a Weyl spinor. What is (χ1 χ2 )† equal to ?

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CHAPTER 3

New Notation for the Components of Weyl Spinors

As mentioned at the end of Chapter 2, you may safely skip this chapter on a ﬁrst reading and come back to it before you tackle Chapter 11. What we would like to do now is to introduce a notation for the components of the spinors that will make it obvious how to construct Lorentz invariants. Our notation will be inspired from the summation convention of relativity, where upper and lower indices can be contracted to form invariant quantities such as Aμ Bμ (or, more generally, to form quantities that have deﬁnite transformation properties, such as tensors). We want our notation to be such that we simply contract the indices of the spinors to form invariants, without having to take hermitian conjugates or to insert iσ 2 matrices. This will force us to introduce more than one type of index, as we will soon ﬁnd out.

44

Supersymmetry Demystified

Since in this section we will be showing the indices of the spinors explicitly in most equations, it would be very awkward to include the L or R labels as well. We will therefore drop these indices with the understanding that χ will always represent a left-chiral spinor, whereas η will consistently be used to represent a right-chiral spinor. When other spinors will be needed, their chirality will be deﬁned when they are introduced. As we have seen previously, we assign lower indices to the components of leftchiral spinors. There is no particular reason for assigning a lower index instead of an upper index and no particular reason to pick left-chiral spinors, but hey, we have to start somewhere! We will use indices taken from the ﬁrst few letters of the Latin alphabet for the spinor components. So let us make the deﬁnition Components of χ ≡ χa What we mean by this is that in all the expressions we have written so far, the χ have to be considered as having lower indices, as is shown explicitly in Eqs. (2.51) and (2.52). Next, we would like to introduce a notation for the components of right-chiral spinors. This is simple if we recall that a Lorentz invariant containing χ and the right-chiral ﬁeld is η† χ . In order to follow the usual convention of Lorentz invariants being built out of matched upper and lower indices, we deﬁne this invariant to be η† χ ≡ ηa χa

(3.1)

You are probably wondering where the hermitian conjugate went! The point is that we choose the deﬁnition (3.1) to make the result as similar as possible to what we encounter in relativity, but the price to pay is that the quantities ηa are not the components of the spinor η themselves but their hermitian conjugate. It’s important to keep this in mind, so let’s emphasize it: ηa ≡ hermitian conjugate of the components of η

(3.2)

The obvious question is then: What notation should we use to represent the components of the spinor η themselves? Well, we would like to keep an upper index, but it has to be different from a Latin letter, and we surely don’t want to introduce a Greek letter, which could cause confusion with four-vector and tensor indices. The solution that has been chosen is to use an upper dotted Latin letter to represent the

CHAPTER 3

New Notation for Weyl Spinors

45

components of a right-chiral spinor, e.g., ηa˙ . This notation in terms of dotted and undotted indices is often referred to as the van der Waerden notation.* Actually, this is not the end of the story. The notation ηa˙ is ﬁne as it is, as long as the indices are shown explicitly. But let’s say that we want to write expressions without including explicit indices. It would be very confusing if, after dropping the indices, we were to write both ηa and ηa˙ as η, so we introduce another layer of notation: We will include a bar symbol over spinors with dotted indices. Our ﬁnal notation for the components of a right-chiral spinor is then Components of η ≡ η¯ a˙

(3.3)

When the indices are shown, the bar notation is redundant, but it is conventional to include the bar whether or not the indices are explicitly written. Again, what we mean by Eq. (3.3) is that in all the equations we wrote before this section, the η has to be thought of as having upper dotted indices. For example, Eq. (3.2) may be written as ηa ≡ (η¯ a˙ )†

(3.4)

η¯ a˙ = (ηa )†

(3.5)

Of course, this also implies

or, to be more explicit, ˙

η¯ 1 = (η1 )†

˙

η¯ 2 = (η2 )†

We see that the effect of taking the hermitian conjugate is to change a dotted index into an undotted index (and to add or remove a bar) without changing the position of the index. Actually, we have only shown this for upper indices on right-chiral spinors, but as we will soon see, it will work for lower indices and for components of left-chiral spinors as well. In case you are wondering, the bar in Eq. (3.3) is indeed the same bar that appears in the spinor dot product in Eq. (2.50), and again, this bar has nothing to do with Dirac conjugation. You may not like this notation at ﬁrst (I sure didn’t) because it may seem to create a potential source of confusion with Dirac conjugation, but *

Bartel Leendert van der Waerden was a Dutch mathematician who made important contributions to several ﬁelds of mathematics, most notably to group theory.

46

Supersymmetry Demystified

there is actually no problem because Dirac conjugation can only be applied to fourcomponent spinors, not to Weyl spinors. In this book we will be working almost exclusively with Weyl spinors, which will always be denoted by lowercase Greek letters. The few four-component spinors we will need will always be denoted by an uppercase . So whenever a bar appears above a Weyl spinor, you will know that it is in the sense of indicating dotted indices, as in Eq. (3.5). So far we have only introduced three types of indexed quantities: left-chiral spinors components with lower undotted indices χa and right-chiral spinors with upper dotted and upper undotted indices η¯ a˙ and ηa . Of our six Lorentz invariants listed in Eq. (2.42), the only one we can recover with our new notation is η† χ, which corresponds to η† χ = (η¯ a˙ )† χa = ηa χa

(3.6)

after using Eq. (3.4). Note that the combination η¯ a˙ χa is not a Lorentz invariant [if it were, we would have η T χ listed in Eq. (2.42)]. This is actually a nice feature that will turn out to be general: in order to obtain a Lorentz invariant, an upper index be matched with a lower index of the same type, i.e., both indices must be dotted or both must be undotted. We clearly need to extend the notation in order to have expressions corresponding to all six invariants of Eq. (2.42). Consider the third invariant in Eq. (2.42): χ † η = (χa )† η¯ a˙

(3.7)

Remember that what we would like is a notation where an upper index must be matched with a lower index of the same type to yield a Lorentz invariant, so it is natural to deﬁne (keeping the tradition that spinors with dotted components have a bar) χ † η ≡ χ¯ a˙ η¯ a˙

(3.8)

Comparing Eqs. (3.7) and (3.8), we see that we must deﬁne χ¯ a˙ ≡ (χa )†

(3.9)

This is a pleasing result because it has the same basic structure as Eq. (3.5): Taking the hermitian conjugate changes a dotted index, into an undotted index, and vice versa, without changing the position of the index.

CHAPTER 3

New Notation for Weyl Spinors

47

To be explicit, Eq. (3.9) corresponds to χ¯ 1˙ = χ1†

χ¯ 2˙ = χ2†

(3.10)

We now have deﬁned left-chiral spinors with both dotted and undotted lower indices as well as right-chiral spinors with the two types of upper indices. What about left-chiral spinors with upper indices or right-chiral spinors with lower indices? To deﬁne those, we could continue with the other invariants in Eq. (2.42), but it is more instructive instead to go back to the Lorentz transformations given in Eq. (2.41). Of course, we want to deﬁne quantities with the same type of indices to have the same Lorentz transformations, so we should deﬁne, e.g., the components χ¯ a˙ to transform the same way as the η¯ a˙ . Looking at Eq. (2.41), we see that it is the quantity iσ 2 χ †T that transforms the same way as a right-chiral spinor, i.e., with the matrix A. We therefore deﬁne χ¯ a˙ ≡ components of iσ 2 χ †T In order for the indices to match, we must write ˙

†

χ¯ a˙ = (iσ 2 )a˙ b χb

(3.11)

or, using Eq. (3.9), ˙

χ¯ a˙ = (iσ 2 )a˙ b χ¯ b˙

(3.12)

where the indices of the iσ 2 were chosen in order to follow the rules that indices must be of the same type to be summed over. Note that in Eq. (3.11) we used a dotted b˙ index on the Pauli matrix even if the spinor had an undotted index because of the presence of the hermitian conjugation, which we know changes an undotted ˙ index into a dotted one. However, some authors write that equation using (iσ 2 )ab , so there is no universally accepted convention in this case. Don’t be alarmed by the freedom in the type of indices assigned to iσ 2 ; in contradistinction with spinor indices, the indices on σ 2 are mostly decorative. We will come back to this point shortly, but for now, just know that no matter what indices we are assigning to it, σ 2 always represents the usual Pauli matrix. To be explicit, Eq. (3.12) gives ˙

χ¯ 1 = χ¯ 2˙

˙

χ¯ 2 = −χ¯ 1˙

(3.13)

Supersymmetry Demystified

48

Using Eqs. (3.10) and (3.13), we may express the components with superscripted dotted indices in terms of the components with subscripted undotted indices: ˙

˙

χ¯ 1 = χ2†

χ¯ 2 = −χ1†

The result [Eq. (3.12)] shows that the matrix iσ 2 may be used to raise dotted indices. We have only shown this for left-chiral spinors, but we will soon prove that this is also valid for right-chiral spinors. Now that we have deﬁned the components of left-chiral spinors with upper dotted indices, the obvious next step is to deﬁne the components χ a . From our earlier discussion, it seems natural to use the deﬁnition χ a = (χ¯ a˙ )† Using Eq. (3.12), we ﬁnd †

˙ χ a = (iσ 2 )a˙ b χ¯ b˙ ˙

= [(iσ 2 )a˙ b ]∗ (χ¯ b˙ )† ˙

where we used the fact that the (iσ 2 )a˙ b are just numbers, so applying a dagger on them amounts to a simple complex conjugation. Now we use Eq. (3.9) to get ˙

†

χ a = [(iσ 2 )a˙ b ]∗ (χb )† = (iσ 2 )ab χb

(3.14)

Since the components of iσ 2 are real, the complex conjugation did not do anything, except that we changed the dotted indices into undotted indices in order, once again, to ensure proper “index etiquette.” As promised earlier, we will come back to the vexing question of the mutating indices of iσ 2 shortly. Note that Eq. (3.14) implies that the matrix iσ 2 may be used to raise undotted indices too. If the derivation of Eq. (3.14) does not satisfy you, we may repeat it in “matrix language” without explicitly showing the indices. We start with χ = (iσ 2 χ) ¯ †T where now we have to specify that the χ on the left has an upper index and the χ¯ has a lower index (we don’t have to specify whether the indices are dotted or

CHAPTER 3

New Notation for Weyl Spinors

49

undotted because the presence or absence of a bar tells us that). We then get χ = (iσ 2 χ¯ )†T = (χ¯ † (iσ 2 )† )T = (χ T (−iσ 2 ))T = (−iσ 2 )T χ = iσ 2 χ It must be understood that the χ on the right has a lower index. This agrees with Eq. (3.14) if we reinstate the indices. The result [Eq. (3.14)] is important enough to deserve its own little box: χ a = (iσ 2 )ab χb

(3.15)

To be explicit, χ 1 = χ2

χ 2 = −χ1

(3.16)

Now that we know how to ﬁnd the components of left-chiral spinors with upper indices in terms of the components with lower indices using Eqs. (3.12) and (3.15), we may invert those two relations to express components with lower indices in terms of components with upper indices. All we need to use is the fact that the inverse of the matrix iσ 2 is −iσ 2 , which is obvious if we recall that (σ 2 )2 = 1. Imposing the indices in the equations to match properly ﬁxes uniquely the indices of −iσ 2 , giving ˙

χ¯ a˙ = (−iσ 2 )a˙ b˙ χ¯ b and χa = (−iσ 2 )ab χ b Therefore, the matrix −iσ 2 is used to lower both dotted and undotted indices. We have only shown this for left-chiral spinors, but we will soon show that it is also valid for right-chiral spinors. So far we have only deﬁned right-chiral spinors with upper indices. If we impose that the ηa transform like the χa and look at Eq. (2.41), we see that the quantity that

Supersymmetry Demystified

50

transforms like χ is −iσ 2 η†T . Thus we deﬁne ˙

ηa ≡ (−iσ 2 )ab (ηb )† = (−iσ 2 )ab ηb As promised, the matrix −iσ 2 is seen to also lower undotted indices on right-chiral † spinors. Deﬁning η¯ a˙ ≡ ηa and following the same steps as in Eq. (3.14), we also can show that ˙

η¯ a˙ = (−iσ 2 )a˙ b˙ η¯ b

(3.17)

which shows that −iσ 2 may be used to lower dotted indices of right-chiral spinors as well. Let us pause to comment on the “chameleon indices” of iσ 2 that change to whatever the equations require. This is something that may be quite confusing at ﬁrst. Note that when we write spinors with different types of indices, e.g., ηa and ηa˙ , we really do mean different quantities. However, it is important to keep in mind that in all our equations, the components of the σ 2 matrix are always the usual components, no matter if the indices are dotted or undotted, no matter if they are upstairs or downstairs. For example, the following components are equal: (iσ )

2 12

2 1˙ 2˙

= (iσ )

=

0

1

−1

0

=1 12

whereas (−iσ 2 )12 = (−iσ 2 )1˙ 2˙ =

0

−1

1

0

= −1 12

It may be confusing to have the matrices with upper and lower indices differing only by a minus sign. If we see the components (iσ 2 )ab in an equation, what prevents us from rewriting this as −(−iσ 2 )ab ? The key point is that we will never deal with σ 2 matrices in isolation; they will always be part of equations containing spinors and other quantities with well-deﬁned Lorentz transformations, and consistency of the equations under those transformations will always clearly dictate what the correct indices are. Let us summarize what we have accomplished so far. We have deﬁned four types of components for both left- and right-chiral spinors. These are all related to one another following three simple rules:

CHAPTER 3

New Notation for Weyl Spinors

51

1. Taking a hermitian conjugate switches dotted and undotted indices without changing their position (and of course adds or removes a bar since a bar always accompanies dotted indices). 2. The matrix iσ 2 may be used to raise indices (both dotted and undotted). 3. The matrix −iσ 2 may be used to lower indices (both dotted and undotted). One last comment before moving on to the topic of Lorentz invariants: It is important to keep in mind that all four sets of components always may be written in terms of a single set and its hermitian conjugate. For example, all types of components of a left-chiral spinor may be written in terms of χ1 and χ2 and their hermitian conjugates χ1† and χ2† , as we have seen explicitly in this section. So despite the fact that we have deﬁned eight different components associated to a single left-chiral spinor, only four of them are independent.

3.1 Building Lorentz Invariants Now let us use our new notation to build Lorentz invariants. As hinted before, the rule to obtain invariants turns out to simply be that an upper index must be matched with a lower index of the same type. We will show that this is the correct rule by recovering all the invariants of Eq. (2.42) this way and nothing else. Following this rule, there are only eight combinations made out of one left-chiral and one right-chiral spinor that can be written: ηa χa

χ a ηa

η¯ a˙ χ¯ a˙

χ¯ a˙ η¯ a˙

χ a χa

χ¯ a˙ χ¯ a˙

ηa ηa

η¯ a˙ η¯ a˙

(3.18)

Of course, reversing the order of any of the spinors does not give a new invariant. For example, χa ηa is simply equal to −ηa χa . In our list (3.18) we chose to list the quantities with undotted indices going diagonally down and dotted indices going diagonally up for a reason that will become clear later on. This does not mean that the quantities with the opposite order are not invariants, just that they are not independent. Still, we seem to get eight invariants, whereas we had found only six in Chapter 2, so our eight expressions must not all be independent. This becomes evident if we recall that we may always write components with upper indices in terms of

Supersymmetry Demystified

52

components with lower indices (without taking a hermitian conjugate) so that, for example, ηa χa and χ a ηa must somehow be related. Indeed, we have ηa χa = (iσ 2 )ab ηb ) χa = −(iσ 2 )ab χa ηb = (iσ 2 )ba χa ηb = χ b ηb

(3.19)

where the minus sign in the second line came from moving the ηb to the right of χa . In the third line, we have used the antisymmetry of iσ 2 to write (iσ 2 )ab = −(iσ 2 )ba , and in the last line, we have used the iσ 2 to raise the index of χ . To summarize, we have shown that ηa χa = χ a ηa

(3.20)

EXERCISE 3.1 If the proof in Eq. (3.19) does not convince you, express explicitly ηa χa in terms of the components with lower indices only (χ1 , χ2 , η1 , and η2 ). Do the same with χ a ηa , and show that the two expressions are equal. Likewise, one can show that η¯ a˙ χ¯ a˙ = χ¯ a˙ η¯ a˙

(3.21)

Therefore, we only have six independent invariants in the list in Eq. (3.18), which we will take to be ηa χa

χ¯ a˙ η¯ a˙

χ a χa

χ¯ a˙ χ¯ a˙

ηa ηa

η¯ a˙ η¯ a˙

(3.22)

It is easy to show that these six invariants correspond indeed to the invariants of Eq. (2.42). The only tricky point is that we must remember that in all the formulas of Chapter 2 the left-chiral spinor χ is always written in terms of its components with lower undotted indices χa , whereas the right-chiral spinor η is always written in terms of its components with upper dotted indices ηa˙ . The claim then is that if we express all the spinors in Eq. (3.22) in terms of these components, we will recover the six expressions of Eq. (2.42).

CHAPTER 3

New Notation for Weyl Spinors

53

For example, consider χ a χa

(3.23)

We need to express χ a in terms of lower indices in order to compare it with Eq. (2.42). We know that the matrix iσ 2 raises indices, so we write

χ a χa = (iσ 2 )ab χb ) χa = χb (iσ 2 )ab χa

(3.24)

where we have used the freedom to move the (iσ 2 )ab around because they are simply numbers. Now, in order to compare with Eq. (2.42), we need to write Eq. (3.24) in matrix form, with the σ 2 matrix sandwiched between the two spinors. To do this, we have to switch the order of the indices of iσ 2 , which brings in a minus sign because of the antisymmetry of iσ 2 . Thus we get χ a χa = χb (−iσ 2 )ba χa = χ T (−iσ 2 ) χ where in the last line it is understood that the indices on the spinors are all lower undotted indices. Sure enough, this is the ﬁrst Lorentz invariant in Eq. (2.42). The other invariants can be obtained in a similar manner. EXERCISE 3.2 Show that η¯ a˙ η¯ a˙ corresponds to one of the invariants of Eq. (2.42). EXERCISE 3.3 At some point we will need the relations equivalent to Eqs. (2.54) and (2.55) but written in terms of upper indices. Prove the two identities 1 χ a χ b = − (iσ 2 )ab χ · χ 2 1 ˙ ˙ χ¯ a˙ χ¯ b = (iσ 2 )a˙ b χ¯ · χ¯ 2

(3.25)

3.2 Index-Free Notation In order to alleviate the notation, it is useful to be able to write expressions without explicit indices by writing, e.g., η χ instead of ηa χa or χ¯ η¯ instead of χ¯ a˙ η¯ a˙ . This is what most SUSY references do, but in this book we will instead use the “spinor

54

Supersymmetry Demystified

dot product notation” introduced in Chapter 2; i.e., we will write η · χ and χ¯ · η. ¯ This makes complex expressions easier to read by showing clearly which spinors are contracted together (e.g., it’s easier to see which spinors are contracted together in χ · ηλ · χ than in χ ηλχ). However, not showing explicit indices obviously leads to an ambiguity. We can distinguish spinors with dotted indices from those with undotted indices because the former have a bar over them, but once we drop indices, we can’t tell if they are supposed to be in the upper or in the lower position. Consider, for example, η · χ . Does this represent ηa χa or ηa χ a ? These two are not equivalent! They differ by a minus sign, as is easily shown: ηa χ a = −χ a ηa = −ηa χa where in the ﬁrst step we have used the Grassmann nature of spin components and in the second step we made use of Eq. (3.20). The same ambiguity arises if we drop the indices of any of the six invariants of Eq. (3.22). We therefore see that if we want to be able to not write explicit indices, we need a new rule to tell us when indices are in the upper position and when they are in the lower position. The convention that has been chosen is that a dot product between unbarred spinors corresponds to the undotted indices written diagonally downward; i.e., η · χ always stands for ηa χa . This convention deserves to be emphasized:

The dot product between two unbarred spinors corresponds to the product of the components with the undotted indices going diagonally downward

What do we do if we want to write ηa χ a as a dot product? We simply need to change the order ﬁrst: ηa χ a = −χ a ηa = −χ · η As for barred spinors, the convention is the reverse: A dot product between barred spinors corresponds to the dotted indices written diagonally upward, i.e.,

CHAPTER 3

New Notation for Weyl Spinors

55

η¯ · χ¯ always stands for η¯ a˙ χ¯ a˙ . Again, let us emphasize this important convention: The dot product between two barred spinors corresponds to the product of the components with the dotted indices going diagonally upward It may seem confusing (and annoying) at ﬁrst that the convention for the barred and unbarred spinors are reversed, but one way to see that it is natural is to remember that we have deﬁned the original components of the left-chiral spinor χ as being χa and the original components of the right-chiral η as being η¯ a˙ . Therefore, in both conventions stated above, the original component of the spinors appears on the right of the spinor dot product. In that sense, the two conventions are consistent with each other. It is easy to check that the dot products we just introduced in terms of components with indices are in fact exactly equal to the dot products introduced in Eq. (2.50). For example, consider η · χ. Using the notation of upper and lower indices, we have η · χ = ηa χa = η1 χ1 + η2 χ2 On the other hand, in the notation of Chapter 2, η · χ = η T (−iσ 2 ) χ = η2 χ1 − η1 χ2 Using Eq. (3.16), we see that these two expressions are indeed equal. Now recall the identities (3.21) and (3.20). Translated into dot-product notation, these become η·χ =χ ·η η¯ · χ¯ = χ¯ · η¯

(3.26)

which we already mentioned in Eq. (2.60) and which you proved in Exercise 2.6. The fact that the order does not matter in a dot product is familiar from relativity

·B

=B

· A.

But keep or even from three-dimensional vector algebra, where A in mind that here the identities (3.26) are more subtle. There is one minus sign introduced by the exchange of the two spinor components, which are Grassmann quantities, and a second minus sign owing to the switch from upper to lower indices on both spinors (or, if you prefer, owing to the antisymmetry of the matrix iσ 2 ). The end result is that we can switch the order of the spinors in those dot products without introducing a minus sign, a result that will come in handy many times.

56

Supersymmetry Demystified

Using dot products, the six Lorentz invariants built out of a left-chiral and a right-chiral spinor of Eq. (3.22) therefore are written as ηa χa = η · χ χ¯ a˙ η¯ a˙ = χ¯ · η¯ χ a χa = χ · χ χ¯ a˙ χ¯ a˙ = χ¯ · χ¯ ηa ηa = η · η η¯ a˙ η¯ a˙ = η¯ · η¯ We now understand the choice of the order of indices in Eq. (3.22). They all have been written with undotted indices going diagonally downward and dotted indices going diagonally upward so that when the time would come to write them as dot products, we would not have to switch the order of anything and introduce minus signs. It’s important to repeat that we could write absolutely everything in this entire book in terms of the components with lower undotted indices and their hermitian conjugates only. Most equations would be more cumbersome because there usually would be a bunch of σ 2 matrices around and hermitian conjugate symbols all over the place, but this is not the main motivation for the introduction of the notation. The key point in using the van der Waerden notation is that it makes it easy to see how expressions transform under Lorentz transformations, which is of major importance in building lagrangians. EXERCISE 3.4 Write down χ · χ χ · η in terms of components with lower undotted indices.

3.3 Invariants Built Out of Two Left-Chiral Spinors In SUSY, it is conventional to work exclusively in terms of left-chiral spinors, so let’s restrict our discussion to this type of spinor from now on. If we have only one such spinor, say, χ , we may write only two invariants: χ · χ and χ¯ · χ. ¯

CHAPTER 3

New Notation for Weyl Spinors

57

If we add a second left-chiral spinor, let’s call it λ, then we now can write six Lorentz invariants: λa χa = λ · χ λ¯ a˙ χ¯ a˙ = λ¯ · χ¯ χ a χa = χ · χ χ¯ a˙ χ¯ a˙ = χ¯ · χ¯ λa λa = λ · λ λ¯ a˙ λ¯ a˙ = λ¯ · λ¯

(3.27)

Again, the identities (3.26) apply: λ·χ =χ ·λ λ¯ · χ¯ = χ¯ · λ¯

3.4 The Notation for ±iσ 2 Writing the matrices ±iσ 2 all the time makes the calculations very awkward, especially when indices are explicitly shown. To avoid this, we will therefore introduce one last layer of notation that is fairly common in SUSY and that consists of using the symbol to denote the matrices ±iσ 2 , following (iσ 2 )ab ≡ ab ˙

˙

(iσ 2 )a˙ b ≡ a˙ b (−iσ 2 )ab ≡ ab (−iσ 2 )a˙ b˙ ≡ a˙ b˙ ˙

We see that ab = a˙ b and ab = a˙ b˙ . It greatly simpliﬁes the notation to use these deﬁnitions, but one must be careful about issues of signs. To be explicit, the notation implies ˙˙

12 = 12 = 1 ˙˙

21 = 21 = −1

Supersymmetry Demystified

58

whereas the expressions with lower indices have the opposite signs: 12 = 1˙ 2˙ = −1 21 = 2˙ 1˙ = 1 Although the with upper indices can be seen as two-dimensional versions of the familiar Levi-Civita symbol i jk , it is important to keep in mind that with lower indices has the opposite sign. Now it is a simple matter to raise or lower indices on spinors using this new notation. We have, for example, χ a = ab χb ˙

χ¯ a˙ = a˙ b˙ χ¯ b

(3.28)

and, of course, similar expressions for the right-chiral spinor. We see that the matrices act as a metric on the spinor indices. However, it is important to keep in mind that in contrast with the metrics of special or general relativity, the are antisymmetric under the exchange of their indices: ab = − ba ˙

˙

a˙ b = − ba˙ ab = −ba a˙ b˙ = −b˙ a˙ In particular, note that the order of the indices in Eq. (3.28) must be respected: The second index of the must match the index of the spinor with which it is contracted; otherwise, a minus sign must be present. For example, ab ψ a = −ba ψ a = −ψb

CHAPTER 3

New Notation for Weyl Spinors

59

In our new notation, Eqs. (2.54), (2.55), and (3.25) take the simpler form

1 ab χ · χ 2 1 χ¯ a˙ χ¯ b˙ = − a˙ b˙ χ¯ · χ¯ 2 1 χ a χ b = − ab χ · χ 2 1 ˙ ˙ χ¯ a˙ χ¯ b = a˙ b χ¯ · χ¯ 2 χa χb =

(3.29)

Obviously, the order in the indices of in Eq. (3.29) also must be respected. Because of this need to pay very close attention to the order of the indices, the use of the is quite a bit more tricky than the use of the metric tensor gμν in general relativity. Still, it is worthwhile to use the notation for the computational simplicity it offers. Since −iσ 2 is the inverse of iσ 2 , we also have that

ab bc = cb ba = δca ˙

˙

ba˙ a˙ b b˙ = δca˙˙ ˙ c = c˙ b˙

(3.30)

where the two δ are the usual Kronecker deltas. We may use the antisymmetry of to generate several new identities, e.g.,

ab bc = − ba bc = − ab cb = ba cb = δca

Note again that the order of the indices is crucial: We get a positive Kronecker delta when the index that is common to the two symbols appears in different positions in the two (ﬁrst and second or second and ﬁrst positions).

60

Supersymmetry Demystified

Let us repeat the proof that η · χ = χ · η [see Eq. (3.19)] to illustrate the power of our new notation: η · χ = ηa χa = ab ηb χa = − ab χa ηb = ba χa ηb = χ b ηb =χ ·η where the ﬁrst minus sign came from the exchange of the two spinor components, and the second minus sign arose from the exchange of the indices.

3.5 Notation for the Indices of σ μ and σ ¯μ Now let’s ﬁnd out what kind of indices we should assign to those two matrices. Recall the results in Eqs. (2.43) and (2.44), which we repeat here for convenience: σ μ i∂μ η transforms like a left-chiral spinor

(3.31)

σ¯ μ i∂μ χ transforms like a right-chiral spinor

(3.32)

whereas

Let’s start with Eq. (3.32). Since we assigned a lower undotted index to χ and we assigned an upper dotted index to quantities transforming like a right-chiral spinor, we see that the matrix σ¯ μ converts a lower undotted index into an upper dotted index, so it must carry two mixed upper indices as follows: ˙ Indices of σ¯ μ = (σ¯ μ )ab

(3.33)

Note that the order is important; the dotted index comes ﬁrst. As for σ μ , Eq. (3.31) tells us that it converts an upper dotted index into a lower undotted index, so Indices of σ μ = (σ μ )a b˙

(3.34)

CHAPTER 3

New Notation for Weyl Spinors

61

We have seen that the matrices may be used to raise or lower indices on the spinors. It is natural to wonder how they may be used to move the indices of σ μ and σ¯ μ . The key relations are provided in Eq. (2.23). Consider the ﬁrst of these identities, which we’ll repeat here: σ 2 σ μ σ 2 = σ¯ μT

(3.35)

According to Eqs. (3.33) and (3.34), σ μ has two lower indices, whereas σ¯ μ has two upper indices. We therefore need matrices to raise the two indices of σ μ ; i.e., we need two matrices iσ 2 . We simply multiply both sides of Eq. (3.35) by i 2 to get iσ 2 σ μ iσ 2 = −σ¯ μT

(3.36)

Let’s take advantage of the notation. Assigning lower indices a b˙ to σ μ , as in Eq. (3.34), we must write the left-hand side of Eq. (3.36) as ˙˙

iσ 2 σ μ iσ 2 = ca (σ μ )a b˙ bd

(3.37)

Note that we do not have any choice on the positions of the indices; since Eq. (3.36) involves the product of three matrices, we must have the second index of the ﬁrst be the same as the ﬁrst index of the matrix σ μ and the second index of the σ μ be equal to the ﬁrst index of the second . And since Eq. (3.37) carries upper indices cd˙ (in that order), we must assign the same indices to the right-hand side of Eq. (3.36). We then obtain that, with all indices explicitly shown, Eq. (3.36) corresponds to ˙

˙˙

ca (σ μ )a b˙ bd = −(σ¯ μT )cd Since a transpose switches the order of indices, we ﬁnally get ˙˙

˙

ca (σ μ )a b˙ bd = −(σ¯ μ )dc Pay close attention to the order of the indices in this expression. This can be written ˙˙ ˙˙ in a more elegant form by using bd = − d b to give us ˙˙

˙

ca d b (σ μ )a b˙ = (σ¯ μ )dc

(3.38)

This is what we were looking for: an expression showing us how the matrices can be used to raise the two indices of σ μ . Note two important points about this identity. First, the indices of σ μ that are raised appear in the second position of both . Second, note how the order of the

Supersymmetry Demystified

62

dotted and undotted indices changes in going from σ μ to σ¯ μ , obeying the deﬁnitions in Eqs. (3.33) and (3.34). The identity (3.38) will be very useful shortly. Of course, we can invert it [or, if you prefer, start from the second relation of Eq. (2.23)] to show that ˙ cb d˙ a˙ (σ¯ μ )ab = (σ μ )cd˙

(3.39)

Let us turn our attention to the ever-important issue of building quantities with deﬁnite Lorentz transformation properties (invariants or tensors). The index assignations (3.33) and (3.34) make it easy to write down such quantities. For example, what contravariant four-vectors can we write down if we have two left-chiral Weyl spinors χ and λ at our disposal? Simply by matching indices, we get the following combinations: ˙ χ¯ a˙ (σ¯ μ )ab λb

˙ χb λ¯ a˙ (σ¯ μ )ab

˙

˙

χ a (σ μ )a b˙ λ¯ b

λa (σ μ )a b˙ χ¯ b

(3.40)

However, only two of those four expressions are independent, as we will demonstrate below. Before doing that, let’s introduce an index-free notation for these expressions. Dropping the indices, we get, for example, ˙ χ¯ a˙ (σ¯ μ )ab λb = χ¯ σ¯ μ λ

(3.41)

As always when we drop indices, some ambiguity might arise. If we were to see the right-hand side appearing in some formula, how would we know if we are to assign upper or lower dotted indices to χ? ¯ The answer is that we have to remember that σ¯ μ carries two upper indices. If we keep this in mind, then it is clear that the right-hand side of Eq. (3.41) corresponds to the expression on the left-hand side. Similarly, we must remember that σ μ carries two lower indices. If we remember this, no ambiguity is possible. We now can rewrite the four terms of Eq. (3.40) unambiguously as χ¯ σ¯ μ λ

λ¯ σ¯ μ χ

χ σ μ λ¯

λ σ μ χ¯

(3.42)

Now let’s show that these expressions are not all independent. Let’s start with the ﬁrst one and use matrices to express the spinors in terms of components with upper indices: ˙ λb χ¯ σ¯ μ λ = χ¯ a˙ (σ¯ μ )ab ˙ = a˙ c˙ χ¯ c˙ (σ¯ μ )ab bd λd

CHAPTER 3

New Notation for Weyl Spinors

63

Now we contract the two with the σ¯ μ using Eq. (3.39) and see where it leads us: ˙ ˙ bd λd = c˙ a˙ db (σ¯ μ )ab χ¯ c˙ λd a˙ c˙ χ¯ c˙ (σ¯ μ )ab

= (σ μ )d c˙ χ¯ c˙ λd

[using Eq. (3.39)]

= −λd (σ μ )d c˙ χ¯ c˙ = −λ σ μ χ¯ where in the ﬁrst step we have switched the indices of both , which produces a factor of (−1)2 = 1. To summarize, we thus have shown that χ¯ σ¯ μ λ = −λ σ μ χ¯

(3.43)

By simply renaming λ ↔ χ , this, of course, also implies χ σ μ λ¯ = −λ¯ σ¯ μ χ

(3.44)

This conﬁrms the claim made earlier that only two of the four expressions in Eq. (3.42) are independent. Let’s now look at what happens if we take the hermitian conjugate of one of these expressions, let’s say χ¯ σ¯ μ λ. Keeping in mind the rule that when we take the hermitian conjugate we switch the order of the components of the spinors without introducing a minus sign, we get (χ¯ σ¯ μ λ)† = λ† (σ¯ μ )† (χ¯ )† = λ† σ¯ μ (χ¯ )†

(3.45)

using the fact that Pauli matrices are hermitian. But since taking the hermitian conjugate on the components of a spinor simply changed undotted indices into dotted indices, the right-hand side of Eq. (3.45) is simply λ¯ σ¯ μ χ. We therefore have established that (χ¯ σ¯ μ λ)† = λ¯ σ¯ μ χ

(3.46)

Similarly, it is easy to show that ¯ † = λ σ μ χ¯ (χ σ μ λ)

(3.47)

Supersymmetry Demystified

64

We’ll close this chapter with the derivation of (yet another!) identity (by the way, all these identities we are proving will be useful at some point, I promise!). This one starts with ¯ σ ν χ) ¯ (λ σ μ χ)(λ

(3.48)

The goal is to rewrite this as something proportional to (λ · λ)(χ¯ · χ¯ ). At ﬁrst sight, this might seem impossible. How can we get the spinors of the same type dotted together when they appear in different matrix products in Eq. (3.48)? What could bring about this kind of magic? The answer lies in the identities in Eq. (3.29). To see how this works, we ﬁrst write Eq. (3.48) in terms of components and move the components of the same type next to one another (recall that we pick up a minus sign when we move a spinor component through another one and that the components of the σ μ matrices are plain numbers): ˙

(λ σ μ χ¯ )(λ σ ν χ¯ ) = λa (σ μ )a b˙ χ¯ b

˙

λc (σ ν )cd˙ χ¯ d ˙

˙

= −(σ μ )a b˙ (σ ν )cd˙ λa λc χ¯ b χ¯ d Now, using Eq. (3.29), we may write this as ˙

˙

−(σ μ )a b˙ (σ ν )cd˙ λa λc χ¯ b χ¯ d =

1 μ ˙˙ (σ )a b˙ (σ ν )cd˙ ac bd λ · λχ¯ · χ¯ 4

(3.49)

Switching the order of the indices in the two matrices and using Eq. (3.38), we get 1 1 μ ˙˙ ˙˙ (σ )a b˙ (σ ν )cd˙ ac bd λ · λχ¯ · χ¯ = ca d b (σ μ )a b˙ (σ ν )cd˙ λ · λχ¯ · χ¯ 4 4 1 ˙ = (σ¯ μ )dc (σ ν )cd˙ λ · λχ¯ · χ¯ 4 By deﬁnition, we have ˙

(σ¯ μ )dc (σ ν )cd˙ = Tr(σ¯ μ σ ν ) This trace turns out to be equal to 2ημν (see Exercise 3.5). Our ﬁnal result therefore is (λ σ μ χ¯ )(λ σ ν χ) ¯ =

1 μν η λ · λχ¯ · χ¯ 2

an identity which will be used several times.

(3.50)

CHAPTER 3

New Notation for Weyl Spinors

65

It’s clear that the same steps also lead to (χ¯ σ¯ μ λ)(χ¯ σ¯ ν λ) =

1 μν η λ · λχ¯ · χ¯ 2

(3.51)

EXERCISE 3.5 Show that Tr(σ¯ μ σ ν ) = 2ημν . Hint: Use Eq. (2.24).

3.6 Quiz 1. Write the invariant η¯ a˙ χ¯ a˙ in an index-free notation (two answers are acceptable). 2. What are 1˙ 1˙ and 1˙ 2˙ equal to? ˙˙

3. What can the expression ac d b (σ μ )a b˙ be simpliﬁed to? 4. What is a˙ b˙ a˙ c˙ equal to? 5. What is the relationship between components with dotted indices and components with undotted indices in the same position? 6. What matrix may be used to lower both dotted and undotted indices? 7. If we contract all the indices in ab cb χ a , what do we get?

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CHAPTER 4

The Physics of Weyl, Majorana, and Dirac Spinors

In Chapter 2 we got acquainted with Weyl spinors and their behavior under Lorentz transformations. However, as noted before, supersymmetric theories are often written in terms of Majorana spinors instead. It is therefore important to understand the connection among Weyl, Majorana, and Dirac spinors both at the mathematical level and at the physical level. In addition, we will see how we can go back and forth between Weyl and Majorana spinors. The ﬁrst step is to understand the operation of charge conjugation applied to spinors and the connection with antiparticle states.

Supersymmetry Demystified

68

4.1 Charge Conjugation and Antiparticles for Dirac Spinors Charge conjugation is the operation that turns a ﬁeld describing a certain particle into a ﬁeld describing the corresponding antiparticle. To study the effect of charge conjugation, let us consider a Dirac spinor because this probably is more familiar to most readers. In this chapter and for the remainder of this book we will drop the L and R labels on the Weyl spinors to alleviate the notation. It will be understood that η will always represent a right-chiral spinor and χ a left-chiral spinor. Consider, then, a Dirac spinor corresponding to a certain particle p (which could be an electron, a neutrino, a quark, etc.):

ηp

p = χp

(4.1)

The antiparticle spinor is obtained by applying the charge-conjugation operator (which we will write down explicitly in an instant) to the particle state, so we may write

cp

η p¯ ≡ p¯ ≡ χ p¯

where the c denotes charge conjugation. Here, a bar over the name of a particle simply will represent the corresponding antiparticle (as in ν¯ for the antineutrino). It is unrelated to the bar used in Chapters 2 and 3. Note that in a Dirac spinor, η p and η p¯ are two different right-chiral spinors. The same comment applies to χ p and χ p¯ ; they are different left-chiral spinors. But clearly, those four spinors are related. To establish the connection, we need to work out explicitly the charge conjugation. The operation of charge conjugation is deﬁned as T

cp ≡ C p

(4.2)

where the bar denotes here the usual Dirac adjoint, and the charge-conjugation matrix C may be taken to have the following representation: C = −iγ γ = 2

0

iσ 2

0

0

−iσ 2

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The Physics of Spinors

69

With a little work, we may write Eq. (4.2) in a more convenient form: T

0 †T †T

cp = C p = C( †p γ 0 )T = C(γ 0 )T †T p = Cγ p ≡ C 0 p

where we have used (γ 0 )T = γ 0 and have introduced C0 = Cγ 0 = −iγ 2 γ 0 γ 0 = −iγ 2 =

0

iσ 2

−iσ 2

0

Here we have used the fact that γ 0 γ 0 = 1. Note that C0 C0 = 1 as well. Written explicitly, the charge-conjugated state is

cp =

†T

iσ 2 χ p

†T

−iσ 2 η p

Using the fact that the charge-conjugated state is the antiparticle state, deﬁned in Eq. (4.1), we ﬁnally ﬁnd η p¯ = iσ 2 χ p†T χ p¯ = −iσ 2 η†T p

(4.3)

This is an interesting result: These expressions are exactly what we obtained in Eqs. (2.40) and (2.37)! What Eq. (4.3) tells us is that the right-chiral antiparticle spinor is the right-chiral quantity built out of the particle left-chiral spinor! Likewise, the left-chiral antiparticle state is given by the left-chiral expression built out of the right-chiral particle spinor. This is, by the way, the reason we introduced factors of i and −i in Eqs. (2.40) and (2.37): it was in order to make the correspondence with charge conjugation exact. It is easy to check that applying charge conjugation twice to a state gives back the initial state: †T ( cp )c = C0 ( cp )†T = C0 (C0 †T = C0 C0∗ p = C02 p = p p )

where we have used that the operation †T applied to a matrix (as opposed to a quantum ﬁeld) is simply a complex conjugation. Applying the charge conjugation operation to both sides of Eq. (4.2), we ﬁnd T

p = C cp

(4.4)

Supersymmetry Demystified

70 which leads to

†T

η p = iσ 2 χ p¯

(4.5)

†T

χ p = −iσ 2 η p¯

Of course, this also can be obtained directly by inverting Eq. (4.3). As expected, the particle left-chiral spinor can be written as the left-chiral quantity built out of the antiparticle right-chiral spinor. Likewise, the particle right-chiral spinor can be written as the right-chiral expression built out of the antiparticle leftchiral spinor. Equations (4.3) and (4.5) teach us an important lesson: We are always free to express the left- and right-chiral states of an antiparticle in terms of, respectively, the right- and left-chiral states of the particle ﬁeld (and vice versa). For example, we may rewrite a Dirac spinor in terms of left-chiral ﬁelds only, at the condition of introducing the left-chiral antiparticle state:

=

ηp χp

=

†T

iσ 2 χ p¯ χp

(4.6)

This point will be crucial when we build the supersymmetric extension of the standard model because we will express all fermions in terms of left-chiral states. It must be kept in mind here that even though only left-chiral ﬁelds appear, the upper two components still transform as a right-chiral spinor, so the four-component spinor always has the structure:

=

quantity that transforms like a right-chiral spinor quantity that transforms like a left-chiral spinor

Of course, we could as well have written the Dirac spinor in terms of right-chiral ﬁelds only:

=

ηp χp

=

ηp †T −iσ 2 η p¯

4.2 CPT Invariance A short but crucial point needs to be mentioned here. Realistic relativistic quantum ﬁeld theories must be CPT-invariant (some authors write TCP or PCT); i.e., they must be invariant under the combined operations of charge conjugation C, parity P,

CHAPTER 4

The Physics of Spinors

χp

CPT

71

ηp¯

Figure 4.1 Two states related by CPT.

and time reversal T. A consequence of this requirement is that lagrangian densities must be “real” in the sense that they must satisfy L† = L

(4.7)

This shows right away why we cannot have a theory containing a left-chiral spinor χ without including as well its hermitian conjugate χ † because a lagrangian containing only one of the two obviously could not respect Eq. (4.7). Now, from Eq. (4.3) we see that χ † can be written in terms of the right-chiral antiparticle spinor η p¯ . This means that the states χ p and η p¯ always appear together in a CPT invariant ﬁeld theory, which we illustrate in Figure 4.1. In physical terms, this implies that if a theory contains a left-chiral spinor, it also necessarily contains the corresponding antiparticle, which will be right-chiral. When promoted to the status of quantum ﬁeld operator, χ p creates the right-chiral antiparticle and annihilates the left-chiral particle. For similar reasons, the pair of ﬁelds shown in Figure 4.2 also must always appear together in a CPT-invariant quantum ﬁeld theory. In other words, if there is a right-chiral particle state, there also must be a corresponding left-chiral antiparticle state. Actually, the two ﬁgures are totally equivalent because it is completely arbitrary what we call the particle and what we call the antiparticle. So we are free to switch the labels p and p¯ in the ﬁrst ﬁgure, which then becomes the second ﬁgure. What we have learned is that if a particle of a certain chirality is present in a theory, there will necessarily be a corresponding antiparticle of opposite chirality. If the particle is massless, chirality and helicity can be used interchangeably. Note that CPT invariance does not constrain us to necessarily have all four states χ p , η p , χ p¯ , and η p¯ present. It is possible to build a completely consistent theory out of χ p and η p¯ alone or out of χ p¯ and η p alone. We will discuss this key observation more fully in the next few sections.

ηp

CPT

χp¯

Figure 4.2 Two other states related by CPT.

Supersymmetry Demystified

72

4.3 A Massless Weyl Spinor Let us start with the simplest fermion theory that we could possibly imagine: the theory of a noninteracting massless two-component Weyl spinor. We will take it to be left-chiral. Then our lagrangian contains only the kinetic term given by Lkin = χ † σ¯ μ i∂μ χ

(4.8)

EXERCISE 4.1 Show that the lagrangian (4.8) is real in the sense of Eq. (4.7). From what we learned in the preceding section, we know that one would observe in the lab a left-chiral particle and its corresponding right-chiral antiparticle (again, we could exchange the particle and antiparticle labels, but we choose to call the left-chiral state the particle). It is instructive to make a count of the degrees of freedom involved. A two-component spinor contains two complex variables, so there are four “off-shell” degrees of freedom (i.e., four degrees of freedom before the equations of motions are imposed). Physical massless spinors must obey the Weyl equations Eq. (2.29), which provide two constraints, leaving two degrees of freedom and therefore two types of observable states. Since we know that the particles observed would be the left-chiral particle and the corresponding right-chiral antiparticle, we should be able to write the lagrangian in † terms of χ p and η p¯ instead of χ p and χ p . Indeed, this can be done using the second relation of Eq. (4.5): †T

χ p = −iσ 2 η p¯ so that

χ p† = η Tp¯ (−iσ 2 )† = η Tp¯ (iσ 2 ) The lagrangian then takes the form Lkin = η Tp¯ iσ 2 σ¯ μ i∂μ χ p Note that this is a completely viable theory; there is no need to include the states η p and χ p¯ . They are not needed to build a consistent theory. Of course, we could choose to add them in, but then they would be totally independent and therefore would correspond to a new particle in its own right.

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73

4.4 Adding a Mass: General Considerations Let’s start again with a left-chiral Weyl spinor χ p , but now we will make it massive. Before proceeding, it’s important to point out that some references restrict the term Weyl spinors to massless spinors, in which case Weyl spinors are always eigenstates of the helicity operator. Others deﬁne Weyl spinors as being the chirality eigenstates, which are irreducible representations of the Lorentz group. With the ﬁrst deﬁnition, it is helicity that is used as the deﬁning property of Weyl spinors, whereas in the second convention, it is behavior under Lorentz transformations that is judged as being more fundamental. For the proponents of the ﬁrst deﬁnition, massive Weyl fermions is an oxymoron, whereas the other camp has no problem with that. Of course, in the massless case, the two schools of thought live in perfect harmony; it is just when discussing massive particles that things get ugly. In this book we will follow the second convention; i.e., we will use the term Weyl spinors to denote the eigenstates of the chirality operator that have the Lorentz transformations given in Eq. (2.32). Therefore, with our convention, it does make sense to talk about massive Weyl spinors. The down side is that our massive Weyl spinors do not have a well-deﬁned helicity, but this is a small price to pay because helicity will play very little role in this book. Let’s get back to the physics of massive fermions. Things are much more interesting than in the massless case. As we argued in Section 2.3, it is possible to use a Lorentz transformation to change the helicity of a massive spinor. This implies that we need both helicities to describe a general massive particle state. From Eqs. (2.30) and (2.31) and the discussion preceding them, we must conclude that if we need both helicity states to describe a massive particle, we need both chiralities as well. Therefore, for a massive fermion, both chiralities of the particle must be present in the theory. We can summarize the situation by saying that if we start with a left-chiral particle state χ p , Lorentz transformations (LT) necessitate the introduction of a right-chiral spinor η p , as illustrated schematically in Figure 4.3. This ﬁgure is not to

χp LT

ηp Figure 4.3 Two states of a massive spin 1/2 fermion required by Lorentz transformations.

Supersymmetry Demystified

74

χp

CPT

ηp¯

LT

ηp Figure 4.4 States associated to a massive left-chiral spinor through Lorentz and CPT transformations.

be interpreted as saying that a Lorentz transformation changes a left-chiral state into a right-chiral state—it is the helicity that is ﬂipped by a Lorentz boost—but as meaning that Lorentz transformations require the existence of both chiral states for a massive fermion. Now comes the interesting part. As we saw in the preceding section, CPT invariance forces us to introduce the charge-conjugate state η p¯ , which is right-chiral. On the other hand, as we have just seen, for a massive particle, Lorentz invariance also forces us to introduce a right-chiral state η p . The situation is summarized in Figure 4.4. The questions that come immediately to mind are: Can we take those two rightchiral states forced on us to be the same? Can we take them to be different physical states? In other words, do we have a choice? The answer to all these questions is yes! We do have a choice, and this leads to deﬁning two different types of four-component spinors.

4.5 Adding a Mass: Dirac Spinors First, let’s consider the case where we choose the two right-chiral states to represent η p¯ . In that case, there are four physical (on-shell) different physical states, η p = distinct states: the left- and right-chiral states of the particle χ p and η p as well as the left- and right-chiral states of the antiparticle χ p¯ and η p¯ . This is illustrated in Figure 4.5. We can combine the left- and right-chiral states of the particle into a fourcomponent spinor that is the familiar Dirac spinor that we all know and love and which we encountered in the ﬁrst chapter:

D =

ηp χp

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75

CPT

χp

ηp¯

LT

LT

ηp

χp¯ CPT

Figure 4.5 The four states of a massive Dirac fermion.

Since we assume the antiparticle states χ p¯ and η p¯ to be distinct from χ p and η p , we have D

Dc = Let us write the mass term for such a Dirac spinor. From the Dirac lagrangian Eq. (2.2), we know that the mass is −m D D = −m † γ0

= −m η†p χ p + χ p† η p

(4.9)

This shows clearly that both χ p and η p are necessary in order to write down a Dirac mass term. As noted earlier, we also could write the Dirac spinor in terms of left-chiral spinors only, using Eq. (4.6). The mass term then is †T −m D D = −m χ pT¯ (−iσ 2 ) χ p + χ p† iσ 2 χ p¯

(4.10)

which, using the spinor dot products deﬁned in Eq. (2.50), can be written in the elegant form −m D D = −m(χ p¯ · χ p + χ¯ p · χ¯ p¯ )

EXERCISE 4.2 Show that the mass term in Eq. (4.11) is real in the sense of Eq. (4.7).

(4.11)

Supersymmetry Demystified

76

4.6 The QED Lagrangian in Terms of Weyl Spinors Later in the book we will be constructing theories directly in terms of Weyl spinors, and it will be very useful to be able to make the connection with the familiar lagrangians of quantum electrodynamics (QED), quantum chromodynamics (QCD), and so on, which are usually all expressed in terms of four-component Dirac spinors. So let’s see what the QED lagrangian looks like when expressed in terms of Weyl spinors. Let’s ﬁrst consider the Dirac lagrangian for an electron (so the subscript p will be replaced by e), which we can obtain directly from Eq. (2.28): LDirac = D γ μ i∂μ − m D

= χe† σ¯ μ i∂μ χe + ηe† σ μ i∂μ ηe − m ηe† χe + χe† ηe

Again, it is useful to write the Dirac spinor in terms of left-chiral spinors only, using Eq. (4.6). This is so because we will later write all the interactions of the standard model and of the minimal supersymmetric standard model in terms of left-chiral spinors. We already took care of the mass term in Eq. (4.11), so let’s concentrate on the kinetic energy: †

i D γ μ ∂μ D = i D γ 0 γ μ ∂μ D =i

χe¯T (−iσ 2 )

0 †

1

1

0

χe

0

σ¯ μ ∂μ

σ μ ∂μ

0

†T

= iχe¯T σ 2 σ μ σ 2 ∂μ χe¯ + iχe† σ¯ μ ∂μ χe

†T

iσ 2 χe¯ χe

(4.12)

which may be written more elegantly as (see Exercise 4.3) †

i D γ μ ∂μ D = iχe¯ σ¯ μ ∂μ χe¯ + iχe† σ¯ μ ∂μ χe

(4.13)

So the Dirac lagrangian is ﬁnally, using Eq. (4.11), †

LDirac = iχe¯ σ¯ μ ∂μ χe¯ + iχe† σ¯ μ ∂μ χe − m(χe¯ · χe + χ¯ e · χ¯ e¯ )

(4.14)

In Chapter 13 we will build the supersymmetric generalization of this lagrangian.

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77

EXERCISE 4.3 Derive Eq. (4.13) from Eq. (4.12). Hint: You will need Eqs. (A.36), and (A.5) and an integration by parts. Let’s now include the coupling of the fermion with the electromagnetic ﬁeld. This is obtained from the Dirac lagrangian by replacing the partial derivative ∂μ by the gauge covariant derivative Dμ deﬁned by Dμ ≡ ∂μ + iq Aμ where Aμ is the gauge (photon) ﬁeld, and q is the electric charge of the particle. For the electron, q = −e (we use the convention that e is positive), so Dμ = ∂μ − ie Aμ In addition, we must add the kinetic energy of the photon, i.e., 1 − Fμν F μν 4 The two new terms coupling the Weyl spinors to the gauge ﬁeld can be obtained from the two terms of Eq. (4.12) by replacing ∂μ by −ie Aμ : †T

Terms containing Aμ = e χe¯T σ 2 σ μ Aμ σ 2 χe¯ + e χe† σ¯ μ Aμ χe †

= −e χe¯ σ¯ μ Aμ χe¯ + e χe† σ¯ μ Aμ χe

(4.15)

where in the second step we have used Eqs. (A.36) and (A.5) in the way described in Exercise 4.3. Note that it would have been incorrect to replace ∂μ by −ie Aμ in Eq. (4.14) (see Exercise 4.4). EXERCISE 4.4 Explain why we would have obtained the wrong coupling of the gauge ﬁeld with the spinors had we replaced ∂μ by −ie Aμ in Eq. (4.14) instead of doing it in Eq. (4.12). Adding Eq. (4.15) to the Dirac lagrangian Eq. (4.14), we ﬁnally obtain the QED lagrangian expressed in terms of left-chiral Weyl spinors: †

†

LQED = iχe¯ σ¯ μ ∂μ χe¯ + iχe† σ¯ μ ∂μ χe − e Aμ χe¯ σ¯ μ χe¯ + e Aμ χe† σ¯ μ χe −m(χe¯ · χe + χ¯ e · χ¯ e¯ ) −

1 Fμν F μν 4

(4.16)

Supersymmetry Demystified

78

There is something quite interesting going on here. In the derivation of the kinetic term of the positron, we have used Eq. (A.36) and an integration by parts. Each operation introduces a minus sign, so the ﬁnal kinetic energy of the positron has the same sign as the kinetic energy of the electron (which did not require any manipulation). On the other hand, in the case of the coupling of the positron to the electromagnetic ﬁeld, we used Eq. (A.36) but, of course, did not perform any integration by parts (there are no derivatives!), so we ended up with the particle and antiparticle spinors having couplings to the electromagnetic ﬁeld of opposite signs. This, of course, was to be expected because these two states have opposite charges! What is interesting is to see how we started from the Dirac lagrangian written in terms of a Dirac spinor, in which there is only one coupling to the electromagnetic ﬁeld, and ended up with an expression containing two couplings to Aμ with opposite signs once we expressed the theory in terms of particle and antiparticle Weyl spinors. We can write Eq. (4.16) even more elegantly by introducing two gauge covariant derivatives acting on the left-chiral particle and antiparticle spinors. Actually, it is customary to use a common symbol Dμ for both derivatives, with the understanding that when it is acting on the electron Weyl spinor χe , it is deﬁned as Dμ χe ≡ (∂μ − ie Aμ )χe

(4.17)

whereas when it is acting on the positron Weyl spinor χe¯ , it carries the opposite charge: Dμ χe¯ ≡ (∂μ + ie Aμ )χe¯

(4.18)

With this convention, we may write Eq. (4.16) as †

LQED = iχe¯ σ¯ μ Dμ χe¯ + iχe† σ¯ μ Dμ χe − m(χe¯ · χe + χ¯ e · χ¯ e¯ ) −

1 Fμν F μν 4 (4.19)

In Chapter 13 we will build the supersymmetric generalization of this lagrangian.

4.7 Adding a Mass: Majorana Spinors Now consider taking the two right-chiral states η p and η p¯ to be one and the same, as illustrated in Figure 4.6. Identifying η p and η p¯ only makes sense if the particle is its own antiparticle. Therefore, this quantum ﬁeld cannot carry quantum numbers that can distinguish a

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The Physics of Spinors CPT

χp

79

ηp¯

LT

ηp Figure 4.6 The two states of a massive Majorana fermion (η p is deﬁned to be equal to η p¯ ).

particle from an antiparticle (e.g., it must have no electric charge or lepton number). In this case, there are only two degrees of freedom, which we can choose to be χ p and η p with η p = η p¯ = iσ 2 χ p†T where we have used Eq. (4.3). It’s easy to check that with this choice, the antiparticle left-chiral state χ p¯ is indeed the same as the particle left-chiral state χ p . Using Eq. (4.5) with the particle and antiparticle labels switched, we ﬁnd χ p¯ ≡ −iσ 2 η†T p †T = −iσ 2 iσ 2 χ p†T = −iσ 2 (iσ 2 )∗ χ p = −iσ 2 (iσ 2 )χ p = χp Instead of expressing everything in terms of χ p and η p , we may, of course, choose to express everything in terms of the left-chiral ﬁeld χ p and its hermitian conjugate † χ p . Note that in terms of degrees of freedom, there is no difference with a massless Weyl spinor. Here, we have all four quantities χ p , η p , η p¯ , and χ p¯ , but they all can

Supersymmetry Demystified

80 †

be expressed in terms of χ p and χ p using η p¯ = η p = iσ 2 χ p†T

χ p¯ = χ p

We now can assemble χ p and η p into a four-component spinor that is called a Majorana spinor M :

M =

ηp χp

=

†T

iσ 2 χ p χp

(4.20)

By construction, a Majorana spinor is unchanged by charge conjugation: c

M = M

which simply expresses the fact that χ p¯ = χ p and η p¯ = η p . In fact, we could have used this as our starting point (and many authors do) in deﬁning a Majorana spinor; i.e., we could have started from a Dirac spinor and imposed the antiparticle states to be the same as the particle states. Let us write down the lagrangian for a free Majorana particle. We insert Eq. (4.20) into the Dirac lagrangian, but we also include an overall factor of 1/2 for reasons that will become clear later: 1 L M = M γ μ i∂μ − m M 2

(4.21)

The theory of a Majorana particle is usually left in this form, but it is instructive to † expand it in terms of the two-component spinors χ p and χ p : 1 † μ 1 χ p σ¯ i∂μ χ p + χ pT (−iσ 2 ) σ μ (iσ 2 ) i∂μ χ p†T 2 2 m T − χ p (−iσ 2 )χ p + χ p† iσ 2 χ p†T 2 1 † μ 1 m = χ p σ¯ i∂μ χ p + χ pT σ¯ μT i∂μ χ p†T − (χ p · χ p + χ¯ p · χ¯ p ) 2 2 2

LM =

(4.22)

where we have again used Eq. (A.5). This expression can be simpliﬁed further using an integration by parts on the second term to write it as 1 1 T μT χ p σ¯ i∂μ χ p†T = − i∂μ χ pT σ¯ μT χ p†T 2 2

(4.23)

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81

This is actually exactly equal to the ﬁrst term of Eq. (4.22), as can be shown using Eq. (A.36): −

1 1 i∂μ χ pT σ¯ μT χ p†T = χ p† σ¯ μ i∂μ χ p 2 2

The Majorana lagrangian (4.22) therefore can be simpliﬁed to L M = χ p† σ¯ μ i∂μ χ p −

m (χ p · χ p + χ¯ p · χ¯ p ) 2

(4.24)

Note the difference between the Majorana lagrangian and the Dirac lagrangian of Eq. (4.14), ignoring the difference of normalization of the mass terms for now. The Dirac lagrangian contains the kinetic terms of two distinct states, χ p and χ p¯ , and it connects them through the mass term. By contrast, the Majorana lagrangian contains a single kinetic term and uses only χ p and its hermitian conjugate to make up the mass term. We see why we had included a factor of 1/2 in front of the kinetic term in Eq. (4.21): This was in order to obtain the canonical normalization after combining the two kinetic terms of Eq. (4.22) into a single term. As for the factor of 1/2 in front of the mass term, it is chosen so that the parameter m appearing in the lagrangian is indeed the mass of the particle, as we will demonstrate in Section 4.9.

4.8 Dirac Spinors and Parity You might wonder why we consider Dirac spinors at all, given that Weyl spinors (or, equivalently, Majorana spinors) are more fundamental and simpler. Why bother with Dirac ﬁelds at all? The answer has to do with parity. Parity transforms χ p and η p into each other. Obviously, a theory containing a single Weyl spinor (or, equivalently, a single Majorana spinor) cannot be parity-invariant. In order to build a parity-invariant theory, such as QED or QCD, one needs both chiral states of the particle, and it is obviously convenient to group these states into a four-component spinor, which turns out to be nothing other than a Dirac spinor. Of course, the CPT-conjugate states are also needed to complete the theory. Consider the Dirac mass term written in the form (4.9) (before expressing η p in † terms of χ p¯ ): −m η†p χ p + χ p† η p

Supersymmetry Demystified

82

This is clearly parity-invariant because it is invariant under the exchange of η and χ . On the other hand, a Majorana mass term obviously violates parity (there is no η p ﬁeld to switch with the χ p !).

4.9 Definition of the Masses of Scalars and Spinor Fields If you are used to working with Dirac spinors, the presence of the factor of 1/2 in the mass terms of the Majorana and Weyl spinors may be somewhat unsettling. Before discussing this issue, let us ﬁrst talk about scalar ﬁelds. The lagrangian of a free real scalar ﬁeld A is LA =

1 μ 1 ∂ A ∂μ A − m 2 A 2 2 2

(4.25)

How do we know that the parameter m is indeed the mass of the particle? By working out the equation of motion of the ﬁeld and comparing it with the KleinGordon equation. For a scalar ﬁeld φ, the Euler-Lagrange equation is ∂μ

∂L ∂L − =0 ∂(∂μ φ) ∂φ

(4.26)

Applying this to Eq. (4.25), we simply get ∂μ ∂ μ A + m 2 A = 0 which is the Klein-Gordon equation for a particle of mass m. The operator ∂μ ∂ μ is often represented by the symbol and is called the d’Alembertian operator. Our result shows that the factor of 1/2 was necessary in Eq. (4.25) for m to be the mass of the particle. Now consider instead a complex scalar ﬁeld φ whose lagrangian is given by Lφ = ∂ μ φ∂μ φ † − m 2 φφ † Note that there are no factors of 1/2 now! We want to check that the m appearing there is indeed the mass of the particle. This time, in applying the Euler-Lagrange

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83

equation, we treat φ and φ † as independent ﬁelds. Applying Eq. (4.26), we get ∂ μ ∂μ φ † + m 2 φ † = 0 which is again the Klein-Gordon for the ﬁeld φ † with mass m. On the other hand, the Euler-Lagrange equation for φ † yields again the Klein-Gordon equation, but this time for the ﬁeld φ, still with mass m. It is clear why the factor of 1/2 is required in the real scalar ﬁeld lagrangian and not in the case of the complex scalar ﬁeld: In the ﬁrst case, the real scalar ﬁeld appears twice in each term of the lagrangian. Let us now turn our attention to spinor ﬁelds. Roughly speaking, a Majorana spinor is to a Dirac ﬁeld what a real scalar ﬁeld is to a complex scalar ﬁeld. Indeed, a Majorana spinor obeys what is essentially a reality condition for spinors, c

M = M . By analogy with scalar ﬁelds, we would expect the free Majorana lagrangian to contain a factor of 1/2 relative to the Dirac lagrangian. This is, of course, a purely hand-waving argument, but we can demonstrate that it is indeed correct by working out the Euler-Lagrange equations, which we write again as applied to the spinor components χa : ∂μ

∂L ∂(∂μ χa )

−

∂L =0 ∂χa

(4.27)

We now apply this to the lagrangian Eq. (4.24) [we will drop the label p on the spinors because there are no antiparticle spinors in Eq. (4.24) and therefore no possibility of confusion]. The only thing we must be careful about is that derivatives with respect to spinor components are Grassmann quantities, so they anticommute with spinors. For example, ∂ ∂ (χ · χ ) = [χb (−iσ 2 )bc χc ] ∂χa ∂χa ∂χc ∂χb (iσ 2 )bc χc + χb (iσ 2 )bc =− ∂χa ∂χa where we picked up a minus sign in the second term by passing the derivative through the spinor component χb . Now we use ∂χb = δba ∂χa

Supersymmetry Demystified

84

where the δ is just the usual Kronecker delta; i.e., δ11 = δ22 = 1, whereas δ12 = δ21 = 0. We then have ∂ χ · χ = δba (−iσ 2 )bc χc + χb (iσ 2 )bc δca ∂χa = (−iσ 2 )ac χc + χb (iσ 2 )ba = (−iσ 2 )ac χc + χb (−iσ 2 )ab = −2(iσ 2 )ac χc

(4.28)

Of course, ∂ χ¯ · χ¯ = 0 ∂χa Using Eq. (4.28), we ﬁnd ∂L = i m (σ 2 )ab χb ∂χa

(4.29)

This takes care of the second term of Eq. (4.27). Now consider the ﬁrst term: ∂μ

∂ ∂(∂μ χa )

† (χ σ¯ i∂ν χ ) = −i ∂μ χb (σ¯ μ )ba †

ν

where the minus sign came from passing the derivative with respect to ∂μ χ to the right of χ † , and we did not show explicit spinor indices to simplify the notation. This may be written as † †T −i ∂μ χb (σ¯ μ )ba = −i (σ¯ μT )ab ∂μ χb

(4.30)

Substituting Eqs. (4.29) and (4.30) into Eq. (4.27), the Euler-Lagrange equation for χ is ﬁnally −i σ¯ μT ∂μ χ †T − i m σ 2 χ = 0 Using the fact that σ 2 σ 2 = 1 and multiplying by i, we may write this as σ¯ μT σ 2 σ 2 ∂μ χ †T + m σ 2 χ = 0

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85

If we multiply this from the left by yet another σ 2 and use Eq. (A.8), we ﬁnally get the Euler-Lagrange equation for χ : σ μ σ 2 ∂μ χ †T + mχ = 0

(4.31)

On the other hand, a similar calculation shows that the Euler-Lagrange equation for χ † is σ 2 σ¯ μ ∂μ χ − m χ †T = 0

(4.32)

If we now isolate χ †T from Eq. (4.32) and substitute it into Eq. (4.31), we ﬁnd σ μ σ¯ ν ∂μ ∂ν χ + m 2 χ = 0

(4.33)

where we again used σ 2 σ 2 = 1. With the help of Eq. (A.9), we may write the ﬁrst term as σ μ σ¯ ν ∂μ ∂ν χ = 2∂ μ ∂μ χ − σ ν σ¯ μ ∂ν ∂μ χ But, after renaming the indices μ ↔ ν in the last term, we get σ μ σ¯ ν ∂μ ∂ν χ = 2∂ μ ∂μ χ − σ μ σ¯ ν ∂ν ∂μ χ which implies σ μ σ¯ ν ∂μ ∂ν χ = ∂ μ ∂μ χ

(4.34)

Using this in Eq. (4.33), we ﬁnally get that χ obeys ∂ μ ∂μ χ + m 2 χ = 0 the Klein-Gordon equation for a particle of mass m. One can easily show that χ † obeys the same equation. This completes the proof that with the factor of one-half included in Eq. (4.24), the parameter m is indeed the mass of the fermion.

4.10 Adding a Mass: Weyl Spinor It is sometimes stated in the literature that it is not possible to write a mass term for a theory containing only a left-chiral particle (together with its right-chiral antiparticle), that it is necessary to introduce a right-chiral particle state. This is

86

Supersymmetry Demystified

incorrect; we can easily write a mass term for a single left-chiral ﬁeld. Actually, we have already done it in Eq. (4.24)! So the lagrangian for a massive left-chiral Weyl fermion is simply LW = χ p† σ¯ μ i∂μ χ p −

m (χ p · χ p + χ¯ p · χ¯ p ) 2

(4.35)

We see that Eq. (4.35) is exactly the same as the Majorana lagrangian expressed in two-component notation given in Eq. (4.24)! There is no physical difference at all between the theory of a massive Weyl fermion and the theory of a Majorana particle. The only difference is a technical one: The Majorana theory is usually written in terms of the four-component Majorana spinor, as in Eq. (4.21), whereas the massive Weyl theory is written in terms of the two-component spinor, as in Eq. (4.35). In other words, we could say that the Majorana theory is simply the four-component version of a massive Weyl fermion (note that the degrees of freedom match; both theories contain two observables states). It’s because of this equivalence that sometimes people refer to Eq. (4.35) as the lagrangian of a Majorana spinor, not of a massive Weyl spinor, keeping the term Weyl spinor for massless spinors. But, as mentioned earlier, in this book we choose to deﬁne Weyl spinors through their Lorentz transformations (2.32), which hold regardless of the mass of the particles. We will reserve the expression Majorana spinor for the four-component spinor satisfying the condition of Eq. (4.7). Let us close this section with an important remark. In the last two sections we chose to express everything in terms of the left-chiral state χ and its hermitian conjugate χ † . It is, of course, possible to build Weyl spinors and Majorana spinors out of η and η† instead, but it is conventional in SUSY to express all fermions in terms of left-chiral states (a choice most likely inﬂuenced by the habit of working with the standard model, in which there are no right-chiral neutrino states). This is why we focused on left-chiral spinors and did not include the corresponding expressions and identities for right-chiral spinors (which, however, can be worked out easily with the information that has been provided).

4.11 Relation Between Weyl Spinors and Majorana Spinors Because of the equivalence between the Majorana and Weyl descriptions, we can always write any expression containing a Weyl spinor as one containing a Majorana spinor and vice versa. Indeed, all supersymmetry can be couched in either Weyl or Majorana language, and many references use only one of the two. Which one

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87

is preferred is obviously a matter of taste, but it could be argued that Weyl spinors are more fundamental because they are the simplest building blocks that can be used to construct supersymmetric theories. The main reason that Majorana spinors nevertheless are used widely is probably because calculations in that language resemble more those made with the familiar Dirac spinors, especially when it comes to working out scattering amplitudes. In this book the emphasis has been put on Weyl spinors, but the main results also will be expressed in the Majorana language, and some calculations of scattering amplitudes will be performed using the Majorana language. In this section we will establish some basic relations between Weyl and Majorana spinors. Let’s start with lagrangian terms without derivatives. We have already seen in Section 4.7 that

M M = χ · χ + χ¯ · χ¯

(4.36)

(we may drop the label p without causing any ambiguity because there is no distinction between the antiparticle and particle states for a Majorana fermion). Keep in mind that the bar over a Majorana spinor denotes the usual Dirac adjoint, †

M = M γ 0. Obviously, we need a second relation if we want to be able to isolate χ · χ and χ¯ · χ¯ in terms of the Majorana spinor. What we are looking for is provided by sandwiching γ 5 between two Majorana spinors. It is easy to check that

M γ 5 M = −χ · χ + χ¯ · χ¯

(4.37)

Inverting Eqs. (4.36) and (4.37), we obtain χ · χ = M PL M χ¯ · χ¯ = M PR M

(4.38)

where the projection operators are deﬁned in Eq. (2.18). Actually, it is straightforward to generalize these relations to products involving distinct spinors. Consider a second left-chiral Weyl spinor λ. Let’s represent the Majorana spinor built out of λ as M . Then the generalizations of Eq. (4.38) are χ · λ = M PL M χ¯ · λ¯ = M PR M Recall that the order in spinor dot products does not matter; i.e., χ · λ = λ · χ and χ¯ · λ¯ = λ¯ · χ¯ . This implies that the order in the corresponding Majorana expressions

Supersymmetry Demystified

88 does not matter either, i.e.,

M PL M = M PL M

M PR M = M PR M which can be checked easily. The next obvious thing to try is to sandwich γ μ and γ5 γ μ between Majorana spinors. One can show that

M γ μ M = χ † σ¯ μ λ − λ† σ¯ μ χ

M γ 5 γ μ M = χ † σ¯ μ λ + λ† σ¯ μ χ which imply χ † σ¯ μ λ = M PR γ μ M λ† σ¯ μ χ = − M PL γ μ M Note that no σ μ matrices appear here. These would appear if we were expressing all the spinors in terms of right-chiral states. Of course, all these relations also can be veriﬁed explicitly by using Eq. (4.20) for a Majorana spinor in terms of Weyl spinors.

4.12 Quiz 1. What is the CPT conjugate state of a right-chiral particle? 2. What is the difference between a Dirac mass term and a Majorana mass term (after they are both expressed in terms of Weyl spinors)? 3. What is the difference between the theory of a Majorana spinor and the theory of a Weyl spinor? 4. Is it possible to write a Dirac spinor in terms of left-chiral ﬁelds only? What about in terms of right-chiral ﬁelds only? 5. Write down the mass term of a left-chiral Weyl spinor. Prove that it is real. 6. What symmetry, is a symmetry of a Dirac spinor but not a symmetry of a Majorana spinor (in the sense that a Dirac spinor has a well-deﬁned transformation under this symmetry, whereas a Majorana spinor doesn’t)?

CHAPTER 5

Building the Simplest Supersymmetric Lagrangian

We are ﬁnally ready to construct our ﬁrst supersymmetric theory! This is where the fun really begins. It took a while to get here, but it was necessary to become familiar with Weyl and Majorana spinors and the notation used to describe them before we could attack supersymmetry (SUSY). This chapter will be short and sweet. The lagrangian considered truly will be the simplest one we could possibly imagine: There will be no mass terms and no interactions. Still, it will be sufﬁcient to learn the essential concepts and to see SUSY in action (no pun intended).

Supersymmetry Demystified

90

5.1 Dimensional Analysis Before building our ﬁrst supersymmetric lagrangian, it will be useful to review how one works out the dimensions of the various quantities we will need. In natural units, both the speed of light and Planck’s constant are set equal to 1. This implies that dimensions of time and length are equal, i.e., [T ] = [L]. It also implies that mass and energy have the same dimensions because E = mc2 becomes E = m in natural units. In addition, since Planck’s constant had dimensions of energy multiplied by time, setting it equal to 1 tells us that [T ] = [L] = [E]−1 . We now can express the units of everything as powers of energy. Instead of writing [E]n for some exponent n, it is simpler to give the exponent. We therefore have the following dimensions for the fundamental scales of time, length, and mass: [L] = [T ] = −1

[M] = 1

Note that this implies that the dimension of a derivative ∂μ is 1. The action S = −i( d 4 xL) is a dimensionless quantity which implies that lagrangian densities (which we often simply call lagrangians) must have dimension 4. Using this, we can work out the dimensions of the ﬁelds we will deal with. For example, the kinetic energy of a scalar ﬁeld is given by ∂μ φ ∂ μ φ

(5.1)

which tells us that a scalar quantum ﬁeld has dimension 1 (because [∂μ ] = 1). The quantum electrodynamic (QED) lagrangian density is given by ¯ μ γ μ − m) − LQED = (P

1 Fμν F μν 4

which tells us that a spinor quantum ﬁeld has dimension 3/2. We worked out the dimension of a Dirac ﬁeld but of course Weyl and Majorana quantum ﬁelds have the same dimension. We also see that the energy-momentum tensor Fμν has dimension 2. Since it is deﬁned as Fμν = ∂μ Aν − ∂ν Aμ we also ﬁnd that the gauge ﬁeld Aμ has dimension 1. The result is the same for nonabelian gauge ﬁelds such as the gluon ﬁelds.

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Building the Langrangian

91

Let us summarize our results: [∂μ ] = [φ] = [Aμ ] = 1 [ D ] = [ M ] = [χ] = [η] =

3 2

(5.2)

5.2 The Transformation of the Fields The simplest supersymmetric model we may consider is composed of two free massless ﬁelds, with one bosonic and the other one fermionic. We therefore consider a Weyl spinor and a complex scalar ﬁeld. By free, we mean that there is no interaction between the two ﬁelds. We could take the Weyl spinor to be either right- or leftchiral, but the convention is to use a left-chiral ﬁeld. Our starting point therefore is the lagrangian L = ∂μ φ ∂ μ φ † + χ † i σ¯ μ ∂μ χ

(5.3)

Now we are ﬁnally ready to introduce the SUSY transformations of the ﬁelds! At this point, we could simply throw at you the transformations of the ﬁelds, and we could check that they leave the lagrangian invariant (up to total derivatives). However, the transformations are quite complicated, and it’s not obvious why anyone would have chosen these particular expressions in the ﬁrst place. Since the goal of this book is to demystify SUSY, we will devote a fair part of this chapter to understanding the logic behind the choice of the supersymmetric transformations. Consider ﬁrst the transformation of the scalar ﬁeld. We consider an inﬁnitesimal transformation proportional to a small parameter ζ . Again, there is no standard notation for this parameter, so always check ﬁrst the convention used when looking at a new reference. It might seem that would be a better choice of notation, but as we saw in Section 3.4, this letter is also used in most SUSY references to represent the matrices ±iσ 2 . In the following calculations, always keep in mind that ζ is inﬁnitesimal, so we will drop terms of order ζ 2 or higher in all our equations. Another important point is that we take ζ to be spacetime-independent; i.e., ∂μ ζ = 0 a relation we will use repeatedly. We are therefore considering global SUSY transformations, as opposed to local transformations, for which ζ would be spacetimedependent (another expression commonly used in the literature for global SUSY is

Supersymmetry Demystified

92

rigid supersymmetry). Making ζ spacetime-dependent—in other words, gauging SUSY—forces one to introduce a gauge ﬁeld that turns out to have the properties of a graviton (the force carrier of gravity), the same way that gauging the global U (1) symmetry of the Dirac lagrangian leads to electromagnetism. For this reason, theories with local SUSY invariance are called supergravity theories (SUGRA for short). However, supergravity necessitates familiarity with general relativity, in particular the tetrad formalism, and is beyond the scope of this book. We postulate that the variation of the scalar ﬁeld is proportional to the Weyl spinor χ (this is, after all, the whole idea of SUSY: Bosonic ﬁelds, are transformed into fermionic ﬁelds, and vice versa). Let us then write, tentatively, φ → φ = φ + δφ

(5.4)

with δφ ζ χ We used the symbol because this relation is still qualitative. We need a bit more work to make this precise. When we write down transformations of ﬁelds, we must ensure two things: the two sides of the equation must have the same dimension, and they must have the same Lorentz transformations. Let’s consider the second requirement ﬁrst. Since φ is a scalar ﬁeld, we must somehow build a Lorentz invariant out of ζ and χ. Since χ is a Weyl spinor, it is clear that we have no choice other than taking ζ to be a Weyl spinor too! This is a key result, so let us emphasize it: The parameter of a global SUSY transformation is a (constant) Weyl spinor What does this mean? It means that ζ is a two-component quantity, namely, ζ1 ζ = ζ2 that takes different values in different frames. We must now decide whether we should take ζ to be a right-chiral or a left-chiral spinor. We do have the choice, but again, the convention is to take it to be left-chiral. This tells us that if we do a Lorentz transformation, ζ transform as χL in Eq. (2.32). Note that ζ is not a quantum ﬁeld, even though it is a spinor, so there is no particle associated to it. Going back to the transformation of φ, we therefore must write down a Lorentz invariant made out of the two left-chiral spinors ζ and χ . But we already know how to do this; we simply take the spinor dot product ζ · χ! Recall that the order does not matter, so we could as well have used χ · ζ . We can ﬁnally write down the

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precise expression for the transformation of φ: δφ = ζ · χ

(5.5)

Now let us turn to the question of the dimensions. Since φ has dimension 1 and χ has dimension 3/2, we obtain that The inﬁnitesimal SUSY parameter ζ has dimension −

1 2

Let’s now consider the transformation of χ. We want it to be linear in ζ and φ or φ † (it would be extremely difﬁcult to build a consistent theory with ﬁelds transforming nonlinearly). We will use φ † for a reason that will become clear in an instant. So we take δχ C ζ φ †

(5.6)

where C is an arbitrary constant that we will have to determine. The reason we did not include any such constant in the transformation of φ is that it could have been reabsorbed into the deﬁnition of ζ . However, once the normalization of ζ is ﬁxed by the transformation of φ, we have no freedom left to get rid of the constant in the transformation of χ, and we will indeed see that C is not equal to 1. Before going any further, we need to check whether the right-hand side of Eq. (5.6) has the same Lorentz transformation properties as the ﬁeld χ (since the variation δχ must, of course, transform the same way as χ itself). In this respect, everything is ﬁne: Since φ is a scalar, the right-hand side of Eq. (5.6) transforms like a left-chiral spinor because ζ is left-chiral. But there is a problem: The left side has dimension 3/2, whereas the right side has dimension 1 − 1/2 = 1/2. So we need to increase the dimension of the right side by one. We do not want to put in another factor of φ because we want the transformation to be linear in the ﬁelds. The only other quantity with dimension 1 at our disposal is ∂μ , so we try δχ C ζ ∂μ φ † Now the dimensions are right, but the Lorentz properties do not match! We need to contract the μ index with something, and it has to be something dimensionless. The only possible terms ﬁtting the bill at our disposal are σ μ and σ¯ μ . Looking at Eq. (2.44), we see that when applied to a left-chiral spinor, it is the matrices σ¯ μ that have well-deﬁned Lorentz transformations. We therefore try δχ C σ¯ μ ζ ∂μ φ †

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but unfortunately, this still doesn’t work! To see the problem and how to ﬁx it, let’s move ∂μ φ † in front (which we can do because it’s not a matrix), δχ C (∂μ φ † ) σ¯ μ ζ

(5.7)

Recall that the quantity σ¯ μ ∂μ χL transforms as a right-chiral spinor [see Eq. (2.44)]. The fact that the derivative is acting on a scalar ﬁeld instead of the spinor does not, obviously, affect the behavior of the expression under a Lorentz transformation, so (∂μ φ † )σ¯ μ ζ also transforms as a right-chiral spinor! Thus the right-hand side of Eq. (5.7) transforms as a right-chiral spinor, whereas what we need is something that transforms as a left-chiral spinor. From Chapter 2, however, we know how to do make something out of a right-chiral spinor that will transform as a left-chiral spinor! The rule is to take the hermitian conjugate and transpose of the right-chiral spinor and then multiply it by −iσ 2 [see Eq. (2.40)]. Applying this to the ﬁrst line of Eq. (5.7), we ﬁnally take the transformation of χ to be: †T δχ = −iσ 2 C (∂μ φ † ) σ¯ μ ζ This has the right dimension and also transforms as a left-chiral spinor. Recall that the operation † applies a hermitian conjugate both in the quantum ﬁeld operator sense and in the matrix sense (a transpose followed by complex conjugation). Obviously, the transpose has no effect on the scalar ﬁeld, so (φ † )†T = (φ † )† = φ The only reason we chose to use φ † in our initial guess for the transformation of the spinor was so that the ﬁnal expression would contain φ. Of course, this is purely a matter of taste; there would have been nothing wrong with having φ † in the ﬁnal result instead! On the other hand, the effect of †T on anything that is not a quantum ﬁeld, including the spinor ζ , is to simply apply a complex conjugation. We therefore get δχ = −iσ 2 C ∗ (∂μ φ) σ¯ μ∗ ζ ∗ = −C ∗ (∂μ φ) iσ 2 σ¯ μ∗ ζ ∗ where ∗

ζ =

ζ1∗ ζ2∗

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Using the fact that σ¯ μ∗ = σ¯ μT and (σ 2 )2 = 1, we may write −C ∗ (∂μ φ) iσ 2 σ¯ μ∗ ζ ∗ = −C ∗ (∂μ φ) iσ 2 σ¯ μT σ 2 σ 2 ζ ∗ which, after using Eq. (A.8), gives our ﬁnal result for the transformation of the spinor: δχ = −C ∗ (∂μ φ) σ μ iσ 2 ζ ∗

(5.8)

5.3 Transformation of the Lagrangian We haven’t even really started yet! All we have done is to write down transformations of the ﬁelds that have the correct dimensions and correct behavior under the Lorentz group. What we want to do now is to see if there is a choice of constant C such that the preceding transformations will leave the action invariant. Recall that the action is the integral of the lagrangian density: S=

d 4 xL

The action is invariant if δS = 0, which implies d 4 xδL = 0 This does not imply that δL itself must be zero. If we assume that the ﬁelds vanish sufﬁciently rapidly at spatial inﬁnity, the action will be invariant if either the variation of the lagrangian density is zero or if it is a total derivative of the ﬁelds: δL = ∂μ F(φ, χ, . . . ) where F is some function of the ﬁelds, and the dots are included because later on we will have other ﬁelds in our lagrangians. Now let us build our ﬁrst supersymmetric theory! Again, the goal is to see if we can ﬁnd a value of the constant C such that the variation of the lagrangian density will be zero after dropping any total derivative. The variation of the lagrangian

Supersymmetry Demystified

96 density Eq. (5.3) is

δL = ∂μ (δφ) ∂ μ φ † + ∂μ φ ∂ μ δ(φ † ) + δ(χ † ) i σ¯ μ ∂μ χ + χ † i σ¯ μ ∂μ δχ = ∂μ (δφ) ∂ μ φ † + ∂μ φ ∂ μ (δφ)† + (δχ)† i σ¯ μ ∂μ χ + χ † i σ¯ μ ∂μ δχ (5.9) where we have used δ(φ † ) = (δφ)† and δ(χ)† = (δχ)† . To see that this is true, let’s take the dagger of Eq. (5.4): φ † → φ † + (δφ)† On the other hand, by deﬁnition, we have φ † → φ † + δ(φ † ) so, clearly, δ(φ † ) = (δφ)† , and we may simply write δφ † without any ambiguity. So the variation of φ † and χ † are simply the hermitian conjugates of Eqs. (5.5) and (5.8). One ﬁnds (see Exercise 5.1): (δφ)† = χ¯ · ζ¯ = χ † (iσ 2 ) ζ ∗

(5.10)

(δχ )† = C (∂μ φ † ) ζ T iσ 2 σ μ

(5.11)

and

EXERCISE 5.1 Prove Eqs. (5.10) and (5.11). Using Eqs. (5.5), (5.8), (5.10), and (5.11) into Eq. (5.9), we get δL = (∂ μ χ † )iσ 2 ζ ∗ ∂μ φ −(∂ μ φ † ) ζ T (iσ 2 ) ∂μ χ I

II

+ C(∂ν φ † ) ζ T (iσ 2 ) σ ν i σ¯ μ ∂μ χ −C ∗ χ † i σ¯ μ σ ν (∂μ ∂ν φ)(iσ 2 ζ ∗ ) III

(5.12)

IV

Note that in all the expressions we will write down, derivatives always will be acting only on the ﬁeld that is immediately to their right. To make this explicit, we will always group together a derivative and the ﬁeld it is acting on inside parentheses. For example, instead of writing ∂μ φχ, which could cause confusion, we will write (∂μ φ)χ .

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Keep in mind that the scalar ﬁeld (or the scalar ﬁeld with any number of derivatives acting on it) can be moved freely through spinors and matrices. Also recall that two lagrangian terms differing by a total derivative are to be considered equivalent. These points are important to remember if you do some calculations and obtain a result that seems at ﬁrst sight to be different from the one given in this book or in any other SUSY reference. In addition, several spinor identities we have seen in previous chapters, the most important of which are listed in Appendix A, may be used to rearrange expressions. It takes some practice to be able to tell quickly whether or not two apparently different expressions are actually equivalent. Coming back to Eq. (5.12), we observe that the ﬁrst and last term contain ζ ∗ , whereas the second and third term contain ζ . Thus each pair of terms must cancel separately (up to total derivatives). Consider the ﬁrst and last terms. At ﬁrst sight, these two terms look very different: The last term contains σ¯ μ and σ ν , whereas these matrices do not appear at all in the ﬁrst term. However, we have already shown how the two are related! Indeed, in Eq. (4.34) we proved that σ μ σ¯ ν ∂μ ∂ν χ = ∂ μ ∂μ χ The proof makes it evident that the identity also holds if the derivatives are acting on a scalar ﬁeld, so we have σ μ σ¯ ν ∂μ ∂ν φ = ∂μ ∂ μ φ = φ

(5.13)

In Section 4.9 we proved this using the identity (A.9). It also can be proven by directly contracting the Lorentz indices, as you are invited to do in Exercise 5.2. EXERCISE 5.2 Prove Eq. (5.13) by expanding explicitly the Lorentz indices (i.e., by writing σ ν ∂ν = σ 0 ∂t + σ i ∂i and so on). Therefore, the sum of the ﬁrst and last terms of Eq. (5.12) is I + IV = (∂ μ χ † )iσ 2 ζ ∗ ∂μ φ − iC ∗ χ † φiσ 2 ζ ∗ Of course, we can move φ around because it is not a matrix quantity and write I + IV = (∂ μ χ † )iσ 2 ζ ∗ ∂μ φ − iC ∗ χ † iσ 2 ζ ∗ φ

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We cannot make those two terms cancel because the derivatives act on different ﬁelds. But we have a trick to move derivatives around: integrations by parts! If we integrate by parts the derivative acting on χ † and use the fact that ∂μ ζ ∗ = 0, we get I + IV = −χ † iσ 2 ζ ∗ φ − iC ∗ χ † iσ 2 ζ ∗ φ

(5.14)

where, as usual, we have dropped the surface term. We see that this vanishes if we take C = −i. Now consider the sum of the second and third terms of Eq. (5.12): II + III = −(∂ μ φ † ) ζ T iσ 2 ∂μ χ + C(∂ν φ † ) ζ T iσ 2 σ ν i σ¯ μ ∂μ χ which, after integrating the derivatives acting on the spinor ﬁeld by parts and using Eq. (5.13), may be written as II + III = (φ † ) ζ T iσ 2 χ − iC (φ † ) ζ T iσ 2 χ

(5.15)

which again vanishes if we take C = −i! We have therefore achieved our goal: We have introduced a SUSY transformation that leaves our action invariant! We have shown that the lagrangian Lfree = ∂μ φ ∂ μ φ † + χ † i σ¯ μ ∂μ χ

(5.16)

is invariant under the SUSY transformations δφ = ζ · χ δφ † = ζ¯ · χ¯ δχ = −i(∂μ φ) σ μ iσ 2 ζ ∗ δχ † = −i(∂μ φ † )ζ T iσ 2 σ μ

(5.17)

Building our ﬁrst supersymmetric theory was not so hard after all! Unfortunately, there is one subtlety that will require us to modify these results signiﬁcantly in two ways: we will need to include a new ﬁeld in the lagrangian (with its own SUSY transformation) and modify the transformation of the spinor. To understand why anyone would want to do that, we will ﬁrst need to review the concept of symmetry charges and their algebra.

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5.4 Quiz 1. What is rigid supersymmetry? 2. What is the dimension of the inﬁnitesimal SUSY parameter ζ ? 3. We want the transformation of the spinor ﬁeld to be linear in the scalar ﬁeld φ and in the SUSY parameter ζ . Using dimensional analysis alone, what form must the variation of the spinor ﬁeld δχ take (the answer does not have to have the correct Lorentz transformation property)? 4. What is SUGRA? 5. What would be wrong with the transformation δχ = C(∂μ φ) σ¯ μ ζ ?

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CHAPTER 6

The Supersymmetric Charges and Their Algebra

When a lagrangian is observed to possess some continuous symmetry, the natural thing to do is to ﬁnd the charges generating the symmetry, also called the generators of the symmetry. Actually, as we will see, what really matters is the algebra obeyed by the charges, i.e., the commutation or anticommutation rules, not the charges themselves (as an aside, when the algebra involves anticommutators, it is often referred to as a graded Lie algebra). We will start by reviewing some elementary results concerning charges and their algebra before tackling the supersymmetry (SUSY) algebra in Section 6.5. We will be led to modify our supersymmetric theory of the preceding chapter substantially in order for the SUSY algebra to be well-deﬁned for off-shell ﬁelds (see Sections 6.6 to 6.8).

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Supersymmetry Demystified

6.1 Charges: General Discussion To explain what charges are, let’s go back to the ﬁeld transformations δφ and δχ , which leave the action invariant. We have written these variations explicitly in terms of the ﬁelds and of the inﬁnitesimal spinor ζ . Now, if we think in terms of canonical quantization, φ and χ are operators, and the calculation of any physical process always boils down to the evaluation of expectation values of some combination of the ﬁelds. To simplify the discussion, consider for now a scalar ﬁeld operator φ (unrelated to the scalar ﬁeld φ that appeared in our supersymmetric lagrangian). If a certain transformation is a symmetry of the theory, it must leave unchanged the expectation value a|φ|b

(6.1)

We will be mostly interested in spacetime transformations and, later, the implementation of the supersymmetric transformations as transformations in superspace. Let us then focus for now on spacetime transformations. Transformations that leave a classic scalar ﬁeld unchanged should leave the expectation value (6.1) unaffected as well. To be speciﬁc, we should have a |φ(x )|b = a|φ(x)|b

(6.2)

Note that there are two differences between the two expressions: The ﬁeld operator is evaluated at different spacetime points, and it is sandwiched between different states. We now introduce a new ﬁeld operator φ that we deﬁne by a |φ(x )|b ≡ a|φ (x )|b

(6.3)

In other words, the effects of the change of states is contained in the deﬁnition of the new ﬁeld operator. If we substitute the right-hand side of Eq. (6.3) back into Eq. (6.2), we get a|φ (x )|b = a|φ(x)|b Since the states are arbitrary, we must conclude that φ (x ) = φ(x) This is not surprising, this is exactly how a classic scalar ﬁeld behaves under a Lorentz transformation!

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Let’s write x as x − δx (the reason for the choice of sign of δx will become clear shortly). This gives us φ (x − δx) = φ(x) or φ (x) = φ(x + δx)

(6.4)

We see that our choice of a negative sign of the variation in x was so that we would get a positive sign on the right-hand side of Eq. (6.4). We will use this in a moment, but ﬁrst, let’s talk about what happens if the ﬁeld is not a scalar. Let’s consider instead a ﬁeld operator F that transforms under some representation of the Lorentz group (a vector, spinor, etc.). Then, Eq. (6.2) will be replaced by a |F(x )|b = S() a|F(x)|b

(6.5)

where S() is the Lorentz transformation matrix corresponding to the representation of the ﬁeld operator. Correspondingly, Eq. (6.4) is replaced by F (x) = S() F(x + δx)

(6.6)

We will need to apply this result to the case of a spinor ﬁeld later in this chapter. Now we will work out a different expression for the left-hand side of Eq. (6.2), an expression that will allow us to establish a connection with the charges. Physically meaningful transformations preserve the norm of states and therefore can be written in the form of unitary operators (or antiunitary operators in the case of time reversal), namely, operators satisfying U † = U −1 . We will take the transformation of the states to be |a → U † |a

|b → U † |b

(6.7)

so the left-hand side of Eq. (6.2) takes the form a |φ(x )|b = a|U φ(x )U † |b It is, of course, arbitrary whether we deﬁne the states to transform with the matrix U or with the matrix U † , and you will encounter both conventions in the literature. Now, using Eq. (6.3), we can relate this expression to the primed ﬁeld φ : a|φ (x )|b = a|U φ(x )U † |b

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Supersymmetry Demystified

which implies φ (x ) = U φ(x )U † or φ (x) = U φ(x)U †

(6.8)

Now we are ready to introduce the charges generating the transformations. Assume that the transformations we are interested in are parametrized by a set of n inﬁnitesimal parameters 1 , 2 , . . . , n , in which case there are necessarily n charges Q 1 , Q 2 , . . . , Q n . The charges are deﬁned through U ≡ exp(±ii Q i ) = exp(±i · Q)

(6.9)

where we introduced a shorthand dot notation for the sum i Q i . We allowed for two possible signs in the argument of the exponential because we will encounter both cases in our calculations. It is important to keep in mind that here the could either be ordinary (commuting) constants or Grassmann quantities, such as the Weyl spinor ζ that we introduced to parametrize supersymmetric transformations. The corresponding charges are then, respectively, bosonic or fermionic. Substituting Eq. (6.9) in Eq. (6.8) and expanding to ﬁrst order in , we get φ (x) ≈ (1 ± i · Q) φ(x) (1 ∓ i · Q) = φ(x) ± i · Q φ(x) ∓ iφ(x) · Q = φ(x) ± i[ · Q, φ(x)]

(6.10)

Let’s deﬁne φ (x) = φ(x) + δφ(x)

(6.11)

By comparing Eqs. (6.10) and (6.11), we obtain a very useful result: ±[ · Q, φ] = −i δφ

(6.12)

where the sign on the left-hand side is the same sign as the argument of the exponential in the operator U . In the case of spacetime symmetries, Eqs. (6.4) and (6.11) imply the obvious relation δφ = φ(x + δx) − φ(x)

(6.13)

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For a general ﬁeld F, we have instead δF(x) = S() F(x + δx) − F(x)

(6.14)

Equations (6.12) and (6.4) [and its generalization, Eq. (6.6)] are the key results of this section. They show us how to relate the transformation of a ﬁeld to a commutator with a charge and to the transformation of the coordinates, respectively. As a very simple example, consider a lagrangian density that is a function of a single real scalar ﬁeld L(φ). The action S = i d 4 xL is obviously invariant under a translation of the ﬁeld φ(x μ ) → φ (x μ ) = φ(x μ + a μ )

(6.15)

where a μ is a constant-displacement four-vector. Therefore, the variation of the ﬁeld δφ under this symmetry transformation is δφ(x μ ) = φ(x μ + a μ ) − φ(x μ ) ≈ a μ ∂μ φ

(6.16)

This is an obvious invariance because the lagrangian density is integrated over all spacetime, so we can always do a shift of the integration variable x μ → x μ + a μ , which implies that d 4 x L(φ) = d 4 x L(φ ). (6.17) Since there are four parameters, the four aμ , we introduce four charges that we suggestively call P μ , and write the unitary operator as U ≡ exp(ia μ Pμ )

(6.18)

Substituting Eqs. (6.16) and (6.18) into Eq. (6.12), we get a μ [Pμ , φ] = −ia μ ∂μ φ

(6.19)

Since this must be true for an arbitrary inﬁnitesimal four-vector aμ , we conclude that [Pμ , φ] = −i∂μ φ a relation that will soon prove useful.

(6.20)

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106

As an aside, the charges of a symmetry sometimes are also called the generators of the transformation. For example, the Pμ are often called the generators of translation in spacetime. As a second simple example, consider a theory containing a single complex scalar ﬁeld and which is described by a lagrangian density that is invariant under a U (1) transformation, φ → eiα φ, with α being a constant parameter. This is an example of an internal symmetry, i.e., a symmetry that is not related to a change of coordinates but to an equivalence between different ﬁelds at the same spacetime point. In this case, the variation of the ﬁeld is δφ = iαφ. Writing U = exp(iα Q), Eq. (6.12) simply gives [Q, φ] = φ

(6.21)

A less trivial example is the object of Exercise 6.1. EXERCISE 6.1 Consider the Lorentz transformation x μ → μν x ν For an inﬁnitesimal transformation, we may write μν x ν ≈ x μ + ωμν x ν = x μ + ωμν xν

(6.22)

where ωμν is antisymmetric, ωμν = −ωνμ . Because of this antisymmetry, it is important to keep track of which index is the ﬁrst one and which is the second one when we write ωμν , which is why the lower index is shifted to the right. We write the unitary operator implementing Lorentz transformation on a scalar ﬁeld as U = e− 2 ω i

μν

Mμν

where the charges M μν are also antisymmetric. The factor of 1/2 is included to allow the sum to be over all values of the indices without double counting (since ω12 M12 = ω21 M21 , for example). Note the sign in the exponent! It is natural to take the Lorentz transformation to have the opposite sign to the translation transformation by analogy with classic physics (see Section 2.5 of Ref. 20 for more details). The transformation of the ﬁeld is given by δφ = φ(x μ + ωμν xν ) − φ(x μ )

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First, prove that we may write 1 δφ = ωμν (xν ∂μ − xμ ∂ν )φ 2

(6.23)

Then prove that the commutation relations between the charges M μν and a scalar ﬁeld are [Mμν , φ] = i(xν ∂μ − xμ ∂ν )φ

The combination of the Lorentz transformations introduced in Exercise 6.1 and the translations generated by the Pμ form the Poincar´e group. You will work out the Poincar´e algebra, the set of commutation relations obeyed by the Lorentz and translation generators, in Exercise 6.3. There is unfortunately no standard in the literature for the signs used in the exponents of the operators U , i.e., for the choice between U and U † in the transformations of the states (6.7) or for the choice between active and passive transformations in the change of coordinates (which affects the sign of δx). A difference of convention in any of these aspects affects the signs of some of the equations containing charges. Because of this, it is typical to see equations differing in signs from one book to the next. I have chosen a set of conventions that maximizes agreement with most introductory references on SUSY, but it’s impossible to agree with all of them because they don’t even agree with one another! Note that Eqs. (6.20) and (6.21) are formal expressions in the sense that they tell us what the commutations of the charges with the scalar ﬁeld are without giving us explicit expressions for those charges. It is possible to obtain explicit representations of the charges, as we will discuss in the next section, but what is even more useful is to work out the algebra of the charges, a topic to which we now turn our attention.

6.2 Explicit Representations of the Charges and the Charge Algebra Given that we already have ﬁgured out the SUSY transformations of the ﬁelds, it might seem like there is no point in ﬁnding the charges. However, the charges associated with a symmetry provide a powerful tool to learn a great deal about theories having this symmetry without even writing down any lagrangian! Actually, we don’t even need to work out the explicit charges; all we really need is the algebra of the charges, i.e., the commutation rules (or anticommutation rules) they obey.

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As an example of how the knowledge of mere commutation rules may be so useful, think about the harmonic oscillator in quantum mechanics. Given only the commutation rules of the raising and lowering operators, it is possible to ﬁnd the allowed energies of the system without even working out any explicit wavefunction! Or consider the spin operators S± and Sz . Using only the algebra between those operators, it is possible to show that particles must have either integer or half-integer spins and that for a given spin s there are 2s + 1 states. In a similar vein, the algebra of the SUSY charges will reveal many general properties of supersymmetric theories, such as the types of particles that must be grouped together in order to form SUSY-invariant theories without us having to write down any lagrangian. Now that we have motivated the usefulness of knowing the algebra of the charges generating a symmetry, let us turn to the question of how to actually determine it. One approach is to build explicit representations for the charges and then to compute by brute force their algebra. It is difﬁcult to use Eq. (6.12) to guess the correct expressions for the charges because of the commutator on the left-hand side. Fortunately, a systematic way to obtain the charges is available. It consists of building a set of currents Jμi associated with the symmetry from which the corresponding charges are found by integrating over space (not spacetime) the zeroth component of the currents: i Q = d 3 x J0i ( x , t) It is important to note that in this approach, the charges (and the currents) are themselves quantum ﬁeld operators. The algebra of the charges then can found using the equal time commutation relations of the ﬁelds present in the theory. We will review the use of currents in detail in Section 7.7. For now, let us simply summarize the strategy underlying this approach. The starting point is a set of ﬁeld transformations leaving the action invariant (i.e., leaving the lagrangian density invariant up to total derivatives). From that, one can work out the algebra of the charges through the following steps: Field transformations ⇒ currents ⇒ charges as quantum ﬁelds ⇒ algebra (6.24) Of course, as a double-check that the charges are correct, one then can use Eq. (6.12) to recover the transformations of the ﬁelds. In the case of spacetime symmetries, there is a second explicit representation of the charges available: as differential operators acting on the ﬁelds. In this chapter we will use carets over the differential operator representation of the charges

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to distinguish them from their representation as quantum ﬁeld operators. Let us now see how these differential operators can be found. The differential operator representation of the charges is deﬁned via ˆ i ) φ(x) φ(x ) ≡ exp(±ii Q ˆ i φ(x) ≈ φ(x) ± ii Q

(6.25)

where now we think of the ﬁeld φ as an ordinary function of spacetime. If we know explicitly the transformation of the coordinates, we simply write x in terms of x and Taylor expand φ(x ) to ﬁrst order in the change of coordinates, which ˆ directly. Once we have differential operators allows us to read off the charges Q representations for the charges, it is a simple matter to compute their algebra. As an example, consider again spacetime translations, Eq. (6.15). Using Pˆ μ to represent the differential operator representation of the charges associated with this symmetry, we write Uˆ = exp(−ia μ Pˆ μ )

(6.26)

Note the sign difference with respect to Eq. (6.18). There is much freedom in the choices of signs in most formula we deal with in this chapter, which makes it quite frustrating to compare equations from different references that typically do not use the same conventions. The absence of standard notation means that the signs of the commutation relations and the signs in the deﬁnitions of the charges will vary from one source to another. This must be kept in mind when consulting the literature. With our convention, we ﬁnd φ(x) + a μ ∂μ φ(x) = φ(x) − ia μ Pˆ μ φ(x)

(6.27)

which leads to Pˆ μ = i∂μ

(6.28)

Note the difference between Eqs. (6.28) and (6.20). It goes beyond the difference of sign on the right-hand side! In Eq. (6.20), the charge P μ is a quantum ﬁeld operator, not a differential operator. EXERCISE 6.2 ˆ μν on a scalar ﬁeld through Deﬁne the action of the differential operators M i μν ˆ

(6.29) ω Mμν φ φ(x ) ≡ exp 2

110

Supersymmetry Demystified

with the change of coordinates given again by Eq. (6.22). Show that ˆ μν = i(xμ ∂ν − xν ∂μ ) M

(6.30)

Be warned that most books do not use a different notation for the charges represented by quantum ﬁeld operators and the charges represented by differential operators, which can lead to some confusion. Even worse, the same symbol is often used to represent the eigenvalues of the charges. For example, consider again P μ . Depending on the context, this may represent a differential operator, a quantum ﬁeld operator, or the actual four-momentum of a state! Always make sure that you know which of these meaning is implied when looking at an equation that contains a charge. In this chapter we use a caret, as in Pˆ μ , for the differential operator representation and a simple capital letter, as in P μ , for the quantum ﬁeld operator. Later, when we will need to represent the eigenvalues of the momentum operator, we will use a lowercase letter, as in pμ . Once we know the representation of the charges as differential operators, it is a trivial matter to work out the algebra they obey (see Exercise 6.3 for the example of the Poincar´e algebra). To summarize, the strategy is to start from transformations of the coordinates that leave the lagrangian density invariant (up to total derivatives) and from there to go through the following steps: Transformation of coordinates ⇒ charges as differential operators ⇒ algebra

(6.31)

At ﬁrst, it may seem as if this approach is irrelevant to SUSY, a symmetry relating boson and fermion ﬁelds. However, as already mentioned in Chapter 1, and as we will show explicitly shortly, there is a deep connection between SUSY and Lorentz transformations, a connection that can be formalized by deﬁning an extension of spacetime called superspace, which contains Grassmann coordinates in addition to the usual spatial and time coordinates. One then can write the SUSY charges as differential operators acting in that extended space. This approach will be pursued in Chapter 10. If the algebra of the charges is known, one actually can reverse the steps illustrated in Eq. (6.31). One may use the algebra to work out the transformation of the coordinates. This is how we will use Eq. (6.31) in Chapter 10, where we will start from the algebra of the SUSY charges to work out the SUSY transformations of the superspace coordinates. We have so far discussed two ways to obtain the algebra of the charges by building explicit representations for them. It turns out that it is also possible to obtain the algebra without ever writing down any explicit representation of the charges. In

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other words, one may start from the transformations of the ﬁelds leaving the action invariant and obtain the algebra directly: Field transformations ⇒ algebra

(6.32)

There is also a shortcut in the case of spacetime symmetries: One can start from the transformation of the coordinates and obtain the algebra of the charges directly: Coordinate transformations ⇒ algebra

(6.33)

In the case of SUSY, the only information we have so far is the ﬁeld transformations that leave the action invariant, so the only two routes available to us to ﬁnd the algebra are Eqs. (6.24) and (6.32) (as noted earlier, the representation of SUSY transformations as changes of coordinates will have to wait until Chapter 10, where we introduce superspace). In the next section we will work out the algebra using the shortcut of Eq. (6.32).

EXERCISE 6.3 Using the representations of the operators P μ and M μν as differential operators [as given in Eqs. (6.28) and (6.30)], show that they obey the so-called Poincar´e algebra, i.e., [ Pˆ μ , Pˆ ν ] = 0 ˆ μν , Pˆ λ ] = i(ηνλ Pˆ μ − ηλμ Pˆ ν ) [M ˆ μσ − ηνσ M ˆ μρ − ημρ M ˆ νσ + ημσ M ˆ νρ ) ˆ μν , M ˆ ρσ ] = i(ηνρ M [M

(6.34)

Warning: Again, several different conventions are used in the literature, which leads to commutation relations differing by an overall sign.

6.3 Finding the Algebra Without the Explicit Charges We will now show how the knowledge of the ﬁeld transformations can be used to obtained the algebra of the charges directly, following Eq. (6.32). In this section, all the charges have to be thought of as quantum ﬁeld operators, not differential operators.

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112

To ﬁnd the algebra of the charges generating a symmetry, we clearly need to consider two transformations in order to obtain an expression with a product of two charges. Let us then consider two successive transformations applied to a ﬁeld φ: a ﬁrst transformation with parameters αi , followed by a second transformation with parameter β j . We will use a dot-product notation for the combinations αi Q i and βj Q j. We have †

Uβ Uα φ Uα† Uβ = exp(iβ · Q) exp(iα · Q) φ exp(−iα · Q) exp(−iβ · Q) ≈ φ + i[β · Q, φ] + i[ α · Q , φ]

− β · Q , [ α · Q , φ] + · · ·

(6.35)

where we neglected all terms of order α 2 or β 2 but kept the term bilinear in α and β. By deﬁnition, Eq. (6.35) is simply a variation with parameter α followed by a variation with parameter β so that we can, of course, also write †

Uβ Uα φ Uα† Uβ = δβ δα φ

(6.36)

We now consider the commutator of two transformations, more precisely δβ δα φ − δα δβ φ

(6.37)

This is equal to

† † Uβ Uα φ Uα† Uβ − Uα Uβ φUβ Uα† ≈ − β · Q , [ α · Q , φ] + α · Q , [ β · Q , φ] (6.38)

Setting Eq. (6.38) equal to Eq. (6.37), we get

δβ δα φ − δα δβ φ = − β · Q , [ α · Q , φ] + α · Q , [ β · Q , φ]

(6.39)

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If we expand this out, being careful about the order of all the quantities involved, we ﬁnd that half the terms cancel out, leaving δβ δα φ − δα δβ φ = α · Q β · Q φ − β · Q α · Q φ − φ α · Q β · Q +φ β · Q α · Q = (α · Q β · Q − β · Q α · Q )φ − φ( α · Q β · Q − β · Q α · Q ) = [ α · Q , β · Q ] φ − φ[ α · Q , β · Q ]

= [ α · Q , β · Q ], φ

(6.40)

The same result, of course, can be obtained using the Jacobi identity, which may be written in the form

A, [B, C] − B, [A, C] = [A, B], C

(6.41)

Applying this to Eq. (6.39) gives right away the last line of Eq. (6.40). The next step is to factor out the inﬁnitesimal parameters, but the details depend on whether the charges are bosonic, which is the case for all continuous symmetries encountered in introductory quantum ﬁeld theory classes, or fermionic, as is the case in SUSY. When the charges are bosonic, Eq. (6.40) leads to the commutation relations between the charges, whereas for fermionic charges, the equation yields anticommutation rules, as we will see explicitly soon.

6.4 Example Before tackling the more complex SUSY case, it might be useful to pause and do a couple of examples with more familiar symmetries. We will illustrate the use of Eq. (6.40) in the case of the algebra between the charges P μ . First, consider the left side of the equation. If we apply a second translation to δa φ(xμ ) with an inﬁnitesimal displacement bμ this time, we get δb δa φ(xμ ) = δb φ(xμ + aμ ) − φ(xμ ) = δb φ(xμ + aμ ) − δb φ(xμ ) = φ(xμ + aμ + bμ ) − φ(xμ + aμ ) − φ(xμ + bμ ) + φ(xμ )

(6.42)

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114

This result is symmetric under the exchange aμ ↔ bμ , which is not surprising because the order of the two translations does not matter. This implies that δb δa φ(xμ ) = δa δb φ(xμ )

(6.43)

so the left-hand side of Eq. (6.40) is zero. Now let us work on the right-hand side of Eq. (6.40). We take the ﬁrst transformation to be Ua = exp(ia μ Pμ )

(6.44)

Ub = exp(ibν Pν )

(6.45)

and the second one to be

In Eq. (6.40) we therefore replace α · Q by a μ Pμ and β · Q by bν Pν , so Eq. (6.40) is ﬁnally given by

[a μ Pμ , bν Pν ], φ = a μ bν [Pμ , Pν ], φ = 0

(6.46)

which implies

[Pμ , Pν ], φ = 0

(6.47)

Now we want to obtain from this the commutator [Pμ , Pν ]. Obviously, [Pμ , Pν ] = 0 is a possible solution, but is it the most general one? This is one difﬁculty in using Eq. (6.40): The ﬁnal step involves extracting from an expression of the form [[Q a , Q b ], φ] the commutator between the charges (or the anticommutator for fermionic charges). This is not always trivial. One must make sure that one writes the most general commutator satisfying Eq. (6.40), and there is, at ﬁrst sight, always more than one possible answer. However, symmetry arguments are usually sufﬁcient to pin down the solution. For example, one could argue that taking the commutator [Pμ , Pν ] to be a constant still would satisfy Eq. (6.47). But we need to have something with two Lorentz indices, so we could try [Pμ , Pν ] = cημν

(6.48)

This obviously satisﬁes Eq. (6.47), but there is a problem: The left-hand side of Eq. (6.48) is antisymmetric under the exchange of the two Lorentz indices, whereas the

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right-hand side is symmetric. This forces us to set c = 0, giving [Pμ , Pν ] = 0

(6.49)

which is indeed the correct answer. The key point of this example is that we obtained the commutation relations of the charges Pμ without ever writing them down explicitly! As a less trivial example, you are invited to prove again the second equation of Eq. (6.34), i.e., [Mμν , Pλ ] = i(ηνλ Pμ − ηλμ Pν ).

(6.50)

EXERCISE 6.4 Prove the commutation relations (6.50) using Eqs. (6.40) and (6.20).

6.5 The SUSY Algebra We will now use Eq. (6.40) to ﬁnd the algebra of the SUSY charges! But ﬁrst let us ask: how many charges do we have? Well, how many inﬁnitesimal parameters are involved in SUSY tranformations? There is the two-component spinor ζ and its complex conjugate ζ ∗ for a total of four parameters. We therefore † † need four charges Q 1 , Q 2 , Q 1 , and Q 2 , which we obviously (see below) can group into a Weyl spinor, that we will simply call Q, and its hermitian conjugate Q † . These SUSY charges are also often referred to as supercharges (what a surprise). Until now, we have used Q to represent charges in general, but for the rest of this book it will be used to speciﬁcally represent the SUSY charges. The argument of the exponential in the unitary operator U must be Lorentzinvariant, so we must combine Q and Q † with ζ and ζ ∗ in a Lorentz-invariant way (which is why it was obvious that the supercharges had to be combined into a Weyl spinor). We choose to make Q a left-chiral spinor. We know from Chapter 2 that the two possible invariant combinations are then Q · ζ = Q (−iσ 2 ) ζ and ¯ · ζ¯ = Q † iσ 2 ζ ∗ Q (recall that ζ is not a quantum ﬁeld operator, so we may write ζ †T simply as ζ ∗ ). Written in terms of the supercharges, the unitary operator U generating SUSY

Supersymmetry Demystified

116 transformations therefore is given by

¯ · ζ¯ ) Uζ = exp(i Q · ζ + i Q

(6.51)

If we apply Eq. (6.12) to the SUSY transformations of φ and χ given in Eq. (5.17), we get ¯ φ] = −iζ · χ [ζ · Q + ζ¯ · Q, ¯ χ] = −i(∂μ φ) σ μ σ 2 ζ ∗ [ζ · Q + ζ¯ · Q, which implies [ζ · Q, φ] = −iζ · χ ¯ χ] = −i(∂μ φ) σ μ σ 2 ζ ∗ [ζ¯ · Q, ¯ φ] = [ζ · Q, χ] = 0 [ζ¯ · Q,

(6.52)

We now consider two successive SUSY transformations of the ﬁelds in order to use Eq. (6.40). We will use β as the inﬁnitesimal parameter of the second transformation. Therefore, the ﬁrst transformation is generated by the operator (6.51), and the second transformation is generated by ¯ · β) ¯ Uβ = exp(i Q · β + i Q Let’s apply Eq. (6.40) to the scalar ﬁeld φ of our supersymmetric lagrangian (we will consider the spinor ﬁeld χ later); we get

¯ · ζ¯ , Q · β + Q ¯ · β], ¯ φ δβ δζ φ − δζ δβ φ = [Q · ζ + Q

¯ · β], ¯ φ = [Q · ζ, Q · β], φ + [Q · ζ, Q

¯ · ζ¯ , Q · β], φ + [ Q ¯ · ζ¯ , Q ¯ · β], ¯ φ + [Q

(6.53)

Right-Hand Side of Eq. (6.53) The right-hand side of Eq. (6.53) can be brought into a much more useful form by writing all the commutators of the charges with the inﬁnitesimal parameters factored out (it will become clear shortly why this is a useful thing to do). Let’s consider the

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ﬁrst commutator, which we rewrite explicitly in terms of the components:

[Q · ζ, Q · β] = Q (−iσ 2 ) ζ, Q(−iσ 2 ) β

= − Q a (σ 2 )ab ζb , Q c (σ 2 )cd βd

(6.54)

where it is understood that the whole expression is acting on the ﬁeld φ. The indices on the σ 2 have been chosen so that the usual summation convention of relativity is respected (if you have read Chapter 3, this should be familiar; if not, don’t read too much into it, it’s just a convention and these σ 2 matrices will drop out of our ﬁnal result anyway). Now we have to be careful. The ζi and βi are Grassmann numbers, but the Q i are Grassmann operators! Thus we can move the βi and ζi through each other and through the Q i on the condition of adding a minus sign, but we cannot switch the order of the Q i (for the same reason that we cannot switch the order of Xˆ and Pˆ in quantum mechanics). As for the components of the σ 2 , these are simple numbers, so we can move them anywhere we wish with no fear. We therefore can write

− Q a (σ 2 )ab ζb , Q c (σ 2 )cd βd = −(σ 2 )ab (σ 2 )cd (Q a ζb Q c βd − Q c βd Q a ζb ) = (σ 2 )ab (σ 2 )cd ζb βd (Q a Q c + Q c Q a ) = (σ 2 )ab (σ 2 )cd ζb βd {Q a , Q c }

(6.55)

Note that we started with a commutator of Q · ζ and Q · β but ended up with an anticommutator of the charges. This is obviously due to the fermionic (Grassmann) nature of the inﬁnitesimal parameters. Equation (6.55) is a useful form because it is written explicitly in terms of the anticommutator of the charges, which will make it easier for us to isolate them in the next section. We still have three commutators to work out in Eq. (6.53). Consider the second one:

¯ · β] ¯ = Q a (−iσ 2 )ab ζb , Q †c (iσ 2 )cd βd∗ [Q · ζ, Q

= Q a (σ 2 )ab ζb , Q †c (σ 2 )cd βd∗ If you have read Chapter 3, you know that, strictly speaking, we should be using dotted indices on the second σ 2 , but again, this will not matter for the present calculation. Now we can simply repeat the steps that led to Eq. (6.55) to get ¯ · β] ¯ = −(σ 2 )ab (σ 2 )cd ζb βd∗ {Q a , Q †c } [Q · ζ, Q

(6.56)

Supersymmetry Demystified

118

The remaining two commutators of Eq. (6.53) are now just a repeat: ¯ · ζ¯ , Q · β] = −(σ 2 )ab (σ 2 )cd ζb∗ βd {Q a† , Q c } [Q ¯ · ζ¯ , Q ¯ · β] ¯ = (σ 2 )ab (σ 2 )cd ζb∗ βd∗ {Q a† , Q †c } [Q

(6.57)

Putting it all together, we see that Eq. (6.53) may be written as δβ δζ φ − δζ δβ φ = [O, φ]

(6.58)

with the operator O being given by the sum of Eqs. (6.55) and (6.56), and the two expressions of Eq. (6.57) O = (σ 2 )ab (σ 2 )cd ζb βd {Q a , Q c } − ζb βd∗ {Q a , Q †c } − ζb∗ βd {Q a† , Q c } + ζb∗ βd∗ {Q a† , Q †c }

(6.59)

Left-Hand Side of Eq. (6.53) Using the SUSY transformations (5.17), we ﬁnd δβ δζ φ = δβ (−ζ T iσ 2 χ ) = −ζ T iσ 2 δβ χ = ζ T iσ 2 iσ μ iσ 2 β ∗ ∂μ φ = −iζ T σ 2 σ μ σ 2 β ∗ ∂μ φ = −iζ T σ¯ μT β ∗ ∂μ φ

using Eq. (A.5)

= iβ † σ¯ μ ζ ∂μ φ

using Eq. (A.36)

(6.60)

where we have used the fact that ∂μ φ is not a matrix, so we are free to move it around. The last line in Eq. (6.60) is clearly the most condensed and most elegant way to write the result, but the form shown two lines above (with the σ 2 still explicit) will be more useful to compute the SUSY algebra because Eq. (6.59) contains σ 2 matrices too. By switching β and ζ , we obviously have δζ δβ φ = −iβ T σ 2 σ μ σ 2 ζ ∗ ∂μ φ = iζ † σ¯ μ β ∂μ φ

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Therefore, the left-hand side of Eq. (6.58) is equal to δβ δζ φ − δζ δβ φ = −i(ζ T σ 2 σ μ σ 2 β ∗ − β T σ 2 σ μ σ 2 ζ ∗ ) ∂μ φ = −i(ζ † σ¯ μ β − β † σ¯ μ ζ )∂μ φ

(6.61)

Again, the ﬁrst form will be the one we will use to obtain the SUSY algebra, but the second one is the most elegant (and will prove useful later on). What we now have to do is to set this equal to the right-hand side of Eq. (6.58). In order to be able to extract explicit expressions for the anticommutators of the SUSY charges, we need to write Eq. (6.61) in the form of the commutator of something with the scalar ﬁeld. The trick is simply to use Eq. (6.20), namely, ∂μ φ = i[Pμ , φ], which allows us to write the ﬁrst line of Eq. (6.61) as

δβ δζ φ − δζ δβ φ = ζ T σ 2 σ μ σ 2 β ∗ − β T σ 2 σ μ σ 2 ζ ∗ Pμ , φ

T 2 μ 2 ∗ T 2 μ 2 ∗ = ζ σ σ σ β − β σ σ σ ζ Pμ , φ

(6.62)

where the last step is possible because ζ , β and the sigma matrices commute with φ. Finally, comparing Eq. (6.62) with Eq. (6.58), we get

[O, φ] = (ζ T σ 2 σ μ σ 2 β ∗ − β T σ 2 σ μ σ 2 ζ ∗ )Pμ , φ

(6.63)

with O given by Eq. (6.59). It is tempting to conclude that this implies O = ζ T σ 2 σ μ σ 2 β ∗ − β T σ 2 σ μ σ 2 ζ ∗ Pμ

(6.64)

but we have only proved Eq. (6.64) when the two sides are applied to the scalar ﬁeld φ. Before concluding that it is a valid identity in our theory, we must prove that it is also satisﬁed when applied to the spinor ﬁeld χ. It turns out, as we will demonstrate in the next section, that Eq. (6.64) is not valid when applied to the spinor ﬁeld. This will lead us to modify our lagrangian to ensure that Eq. (6.64) does apply to all ﬁelds. Since the end result is that Eq. (6.64) will be satisﬁed, let’s go ahead and use it to work out the SUSY algebra. The right-hand side of Eq. (6.64) is not yet in a suitable form for comparison with Eq. (6.59) because we still need to factor out the inﬁnitesimal parameters and write explicit indices on everything. To facilitate the comparison, we will assign the same indices to ζ and β as they have in Eq. (6.59), i.e., b for ζ and d for β. We

Supersymmetry Demystified

120 therefore write

ζ T σ 2 σ μ σ 2 β ∗ = ζb (σ 2 )ba (σ μ )ac (σ 2 )cd βd∗

(6.65)

β T σ 2 σ μ σ 2 ζ ∗ = βd (σ 2 )dc (σ μ )ca (σ 2 )ab ζb∗

(6.66)

and

Note that in Eq. (6.59) the indices of the two σ 2 matrices are, respectively, ab and cd. We would like to have the same indices in our two σ 2 matrices in Eqs. (6.65) and (6.66) because then we will be able to cancel them out in Eq. (6.64). This is easy to achieve because σ 2 is antisymmetric, so (σ 2 )ba = −(σ 2 )ab and (σ 2 )dc = −(σ 2 )cd . In addition, we will move the ζb∗ to the left of the βd in Eq. (6.66) because this is the order in which these appear in Eq. (6.59). Using the fact that the components of the σ matrices are simply numbers that can be moved around freely, we can rewrite Eq. (6.65) as ζ T σ 2 σ μ σ 2 β ∗ = −ζb βd∗ (σ 2 )ab (σ 2 )cd (σ μ )ac

(6.67)

and Eq. (6.66) as β T σ 2 σ μ σ 2 ζ ∗ = ζb∗ βd (σ 2 )ab (σ 2 )cd (σ μ )ca Using these last two results in Eq. (6.64), we obtain O = −(σ ) (σ ) 2 ab

2 cd

ζb βd∗ (σ μ )ac

+

ζb∗ βd (σ μ )ca

Pμ

(6.68)

Using Eq. (6.59) for O and canceling out the components (σ 2 )ab (σ 2 )cd appearing on both sides, we ﬁnally get ζb βd {Q a , Q c } − ζb βd∗ {Q a , Q †c } − ζb∗ βd {Q a† , Q c } + ζb∗ βd∗ {Q a† , Q †c } = −ζb βd∗ (σ μ )ac Pμ − ζb∗ βd (σ μ )ca Pμ

(6.69)

Since the spinors ζ, β and their complex conjugates are arbitrary, Eq. (6.69) actually contains four independent equations. The terms in ζb βd and in ζb∗ βd∗ lead to {Q a , Q b } = 0

(6.70)

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121

and †

{Q a† , Q b } = 0

(6.71)

If you have read Chapter 3, you know that using the dotted and bar notation, this also can be written as ¯ b˙ } = 0 ¯ a˙ , Q {Q Now, if we consider the terms in ζb βd∗ in Eq. (6.69), we ﬁnd {Q a , Q †c } = (σ μ )ac Pμ which is usually written with the index on Q † labeled b: †

{Q a , Q b } = (σ μ )ab Pμ

(6.72)

Again, using the bar and dotted notation of Chapter 3, this may be written as ¯ b˙ } = (σ μ )a b˙ Pμ {Q a , Q

(6.73)

It is important to point out that the normalization of the supercharges is arbitrary, so we are free to rescale Eqs. (6.72) and (6.73) by a constant. Another way to put it is that we could have multiplied the supercharges in the exponential of Uζ by an arbitrary factor. A normalization that is used often √ in the SUSY literature corresponds to rescaling our supercharges by Q → Q/ 2, which then transforms Eqs. (6.72) and (6.73) into †

{Q a , Q b } = 2(σ μ )ab Pμ ¯ b˙ } = 2(σ μ )a b˙ Pμ {Q a , Q

(6.74)

We will use the normalization Eq. (6.72). Let’s ﬁnish with the terms in ζb∗ βd in Eq. (6.69). These yield the relation {Q a† , Q c } = (σ μ )ca Pμ

(6.75)

It is easy to see, however, that this is equivalent to Eq. (6.72). By the deﬁnition of † an anticommutator, we may exchange Q a and Q c (without picking up any minus

Supersymmetry Demystified

122 sign) and write Eq. (6.75) as

{Q c , Q a† } = (σ μ )ca Pμ which, after renaming the indices, is exactly the same as Eq. (6.72). The commutation relations (6.70), (6.71), and (6.72) are what we were seeking: the algebra of the SUSY charges. The fact that the anticommutators (6.72) contain the four-momentum operator is a crucial aspect of supersymmetry, and we will discuss this in more detail shortly. To complete the algebra, we need to work out the commutators of the SUSY charges with the Poincar´e generators P μ and M μν . It is easy to see that the supercharges commute with the momentum operator because the SUSY transformations do not generate any function of x on which the derivatives of P could act, so

Q a , Pμ = 0

† Q a , Pμ = 0

(6.76)

Things are a bit more interesting when it comes to the commutator with the Lorentz generators Mμν . To work this out, we ﬁrst need to write down the Lorentz transformation of Weyl spinors. The Lorentz transformation of a spinor ﬁeld has two contributions: The argument transforms in the same way as for a scalar ﬁeld [see Eq. (6.23)], but in addition, the spinor gets multiplied by the matrices shown in Eq. (2.32). Our ﬁrst step is to write down a covariant expression for the transformation of the spinors given in Eq. (2.32) in terms of the parameters ωμν . For a left-chiral Weyl spinor, the answer is that Eq. (2.32) may be written as

i μν χ = exp ω σμν χ 2

(6.77)

where the matrices σμν are deﬁned as i σμν ≡ (σμ σ¯ ν − σν σ¯ μ ) 4

(6.78)

Note that σμν is antisymmetric under the exchange of the indices μ and ν. The exponential can be shown to reproduce the transformation of χ given in Eq. (2.32), with an appropriate choice of the coefﬁcients ωμν .

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As for a right-chiral spinor η, it transforms as in Eq. (6.77) but with σμν replaced by i σ¯ μν ≡ (σ¯ μ σν − σ¯ ν σμ ) 4

(6.79)

The complete Lorentz transformation of a left-chiral spinor therefore is

i μν χ → χ (x) = exp ω σμν 2

χ (x μ + ωμν xν )

1 i ≈ χ + ωμν σμν χ + ωμν xν ∂μ χ − xμ ∂ν χ 2 2 1 = χ + ωμν iσμν χ + xν ∂μ χ − xμ ∂ν χ 2

(6.80)

so the variation of the spinor is given by 1 δω χ = ωμν iσμν χ + xν ∂μ χ − xμ ∂ν χ 2

(6.81)

To work out the commutation relation of the supercharges Q a with the Lorentz generators Mμν , we start again from Eq. (6.40), which now takes the form −

1

[ζ · Q, ωμν Mμν ], φ = δω δζ φ − δζ δω φ 2

(6.82)

where the factor of −1/2 is the one that accompanies the generators Mμν (see Exercise 6.1). It is then a simple matter to show that [Q a , Mμν ] = (σμν )ab Q b

(6.83)

the proof of which is the subject of Exercise 6.5. The indices of σμν are chosen this way in order to satisfy the usual summation convention. The matrices σμν are antisymmetric, which is why we need to use spaces to indicate clearly that the upper index is the second index. The only tricky aspect of the derivation is that we need at some point the commutator of the supercharges with the scalar ﬁeld. This is easy to ﬁnd by going back to Eq. (6.12), which, using the SUSY transformation of the scalar ﬁeld δφ = ζ · χ,

Supersymmetry Demystified

124 becomes

[ζ · Q, φ] = −iζ · χ

(6.84)

This immediately gives us [Q a , φ] = −iχa or, suppressing the spinor indices, [Q, φ] = −iχ

(6.85)

Using this result, it is a simple matter to prove Eq. (6.83) starting from Eq. (6.82). This calculation is the object of the ﬁrst part of Exercise 6.5. † The calculation of the commutation relation with the charges Q a and Q a† is much simpler using the van der Waerden notation of Chapter 3. We will simply quote the result and leave the derivation to the second part of Exercise 6.5. If you haven’t covered Chapter 3 yet, you may skip this exercise for now and come back to it later. The ﬁnal results are

¯ a˙ , Mμν = − Q ¯ b˙ (σ¯ μν )b˙a˙ Q

a˙ ¯ , Mμν = (σ¯ μν )a˙˙ Q ¯ b˙ Q b

(6.86)

We have achieved our goal of obtaining the complete algebra of the supercharges. However, as mentioned earlier, we have cheated in the step from Eq. (6.63) to Eq. (6.64) because we have not shown that Eq. (6.64) was valid when applied to the spinor ﬁeld as well as to the scalar ﬁeld. We will remedy this lapse in the next section. EXERCISE 6.5 a. Prove the commutation relation (6.83). b. Prove Eq. (6.86) to be done after you have gone through Chapter 3. ¯ which are related to Hint: In order to get an expression for the supercharges Q, the hermitian conjugates of the supercharges Q, we use Eq. (6.40), but with the transformations on the left side applied to φ † instead of φ. So the starting point

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125

will be −

1 ¯ ¯ μν [ Q · ζ , ω Mμν ], φ † = δω δζ φ † − δζ δω φ † 2

(6.87)

It will then be straightforward to prove the ﬁrst commutator of Eq. (6.86). You can then use symbols to obtain the second commutator from the ﬁrst. You will need to show that ˙

c˙ b d˙ a˙ (σ¯ μν )a˙ b˙ = (σ¯ μν )c˙ d˙

(6.88)

6.6 Nonclosure of the Algebra for the Spinor Field As mentioned earlier, the SUSY algebra we have just obtained is, strictly speaking, valid only when applied to the scalar ﬁeld. To prove that the algebra is valid in general, we have to check if we would have obtained the same result starting from the transformation of the spinor ﬁeld. We will now see that it will not work without modifying the theory in a nontrivial way. Consider the variation of the spinor ﬁeld. Using Eq. (5.17), we get δβ δζ χ = δβ (−iσ μ iσ 2 ζ ∗ ∂μ φ) = (σ μ σ 2 ζ ∗ ) ∂μ (δβ φ) = (σ μ σ 2 ζ ∗ ) β · ∂μ χ where, obviously, β · ∂μ χ stands for the usual spinor dot product between β and ∂μ χ: β · ∂μ χ = β T (−iσ 2 )∂μ χ Note that what we have is really two identities, one for each component of χ. To make this more clear, we can write δβ δζ χa = (σ μ σ 2 ζ ∗ )a β · ∂μ χ

(6.89)

Supersymmetry Demystified

126 To simplify the notation, let’s deﬁne

α ≡ σ μσ 2ζ ∗

(6.90)

δβ δζ χa = αa β · ∂μ χ

(6.91)

so that we may write Eq. (6.89) as

Our ultimate goal is to check if the commutator of two SUSY transformations on the spinor is of the same form as what we obtained for the scalar ﬁeld in Eq. (6.61). To compare, we need to rewrite Eq. (6.91) in the form of something times ∂μ χa , i.e., we want the index a to be on χ. We can move the index using the following identity, valid for any three spinors α, β and γ : αa (β · γ ) + βa (γ · α) + γa (α · β) = 0

(6.92)

Note that the spinor dot products are commuting quantities, so we are free to move them around. For example, αa (β · γ ) = (β · γ )αa . EXERCISE 6.6 Prove Eq. (6.92). Hint: Expand out all the terms in terms of the components of the spinors, and check explicitly that the identity is satisﬁed for both a = 1 and a = 2. Obviously, to apply Eq. (6.92) to Eq. (6.91), we simply have to deﬁne ∂μ χ ≡ γ . We then obtain [using Eq. (6.90)] δβ δζ χa = −(∂μ χ ) · α βa − α · β ∂μ χa = −(∂μ χ ) · α βa − β · α ∂μ χa = − (∂μ χ )T (−iσ 2 )σ μ σ 2 ζ ∗ βa − β T (−iσ 2 )σ μ σ 2 ζ ∗ ∂μ χa = i (∂μ χ )T σ 2 σ μ σ 2 ζ ∗ βa + i β T σ 2 σ μ σ 2 ζ ∗ ∂μ χa where in the second line we have rewritten α · β as β · α only so that the transpose would be applied to β, not to the complicated α [deﬁned in Eq. (6.90)]. Note that the expressions inside the two parentheses in the last line are commuting quantities. We can freely move βa or ∂μ χa through these parentheses without picking up any minus sign.

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Using Eq. (A.5) and then Eq. (A.36), we can simplify the result greatly: δβ δζ χa = i (∂μ χ )T σ 2 σ μ σ 2 ζ ∗ βa + i β T σ 2 σ μ σ 2 ζ ∗ ∂μ χa = i (∂μ χ )T σ¯ μT ζ ∗ βa + i β T σ¯ μT ζ ∗ ∂μ χa = −i ζ † σ¯ μ ∂μ χ βa − i ζ † σ¯ μ β ∂μ χa The second term on the right-hand side is what we wanted, something proportional to ∂μ χa , but the ﬁrst term will cause us some grief. Obviously, by switching ζ and β, we get δζ δβ χa = −i β † σ¯ μ ∂μ χ ζa − i β † σ¯ μ ζ ∂μ χa so that

δβ δζ − δζ δβ χa = −i ζ † σ¯ μ ∂μ χ βa + i β † σ¯ μ ∂μ χ ζa − i ζ † σ¯ μ β − β † σ¯ μ ζ ∂μ χa

(6.93)

Note that the last term is identical to the transformation we had for the scalar ﬁeld (6.61) with φ replaced by χ! Unfortunately, the ﬁrst two terms of Eq. (6.93) spoil the fun. Note that if i σ¯ μ ∂μ χ = 0, the two unwanted terms drop out. But this is nothing other than the Weyl equation! In other words, if the spinor satisﬁes its equation of motion (i.e., if it is on-shell), then we recover the algebra we had obtained previously using the transformations of the scalar ﬁeld. However, for an off-shell spinor, one cannot write the algebra of the SUSY and Poincar´e charges in terms of those same charges, as we did in Section 6.5. The conclusion therefore is that the algebra on the spinor ﬁeld closes only on-shell. What we would like to do now is to modify the theory so that the algebra on the spinor ﬁeld will close off-shell, i.e., even if the spinor does not obey its equations of motion. Why do we want that? This is a subtle issue. There are SUSY theories that have no off-shell formulations, i.e., whose algebra closes only on-shell, so it would not be the end of the world if we would not modify our toy theory to close off-shell. However, when the algebra does not close offshell, SUSY transformations become nonlinear (they contain products of the ﬁelds) and depend on the coupling constants present in the theory. This makes it much more difﬁcult to design lagrangians that are invariant under SUSY. We will therefore modify the theory to enforce that the algebra closes off-shell.

128

Supersymmetry Demystified

6.7 Introduction of an Auxiliary Field There is a simple argument requiring no calculations that we could have used to predict that the SUSY algebra would not close off-shell. SUSY requires the number of bosonic and fermionic degrees of freedom to be equal (this will be demonstrated in the next chapter). A Weyl spinor has two complex components and so four degrees of freedom when it is off-shell. On-shell, the equation of motion imposes two constraints, reducing the number of degrees of freedom to two. A complex scalar ﬁeld has two degrees of freedom, so we see that on-shell, the number of bosonic and fermionic degrees of freedom match. However, off-shell, the number of bosonic and fermionic degrees of freedom do not match anymore. This explains why the SUSY algebra only closes on-shell if we have a single complex scalar ﬁeld and a Weyl spinor. But the argument also suggests a solution. We need to add a bosonic ﬁeld that will provide two degrees of freedom off-shell but no on-shell degrees of freedom! A ﬁeld that has no on-shell degrees of freedom is called an auxiliary ﬁeld, so what we need is to add to the theory a complex scalar auxiliary ﬁeld, commonly called F. How do we get the ﬁeld F to have no on-shell degrees of freedom? The trick is to make sure that the equation of motion for this ﬁeld is F † = F = 0! This will clearly be the case if we add a term in the lagrangian that contains a product of F and F † ; i.e., we do not include a kinetic term for F. The simplest real term depending on F and F † is FF† . So our lagrangian now takes the form L = ∂μ φ∂ μ φ † + χ † i σ¯ μ ∂μ χ + FF†

(6.94)

Note that despite being a scalar ﬁeld, the ﬁeld F is of dimension 2! This is something important to remember, so let’s emphasize it: [F] = 2 What should we take for the transformation of F under SUSY? Since it has a different dimension than φ, we cannot use the same transformation. We want a transformation that is linear in the SUSY parameter ζ and linear in one of the ﬁelds. Recall that ζ has dimension −1/2, so it must be multiplied by something of dimension 5/2 (because F has a dimension equal to 2). This means that the transformation must contain the spinor ﬁeld χ, so our starting point is δF ζχ

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Since the dimensions of χ and ζ add up to 1, we need to include something with a dimension equal to 1. We don’t want to use the scalar ﬁeld to avoid a nonlinear transformation. What else do we have with dimension equal to 1? A derivative, of course! So let’s try δ F ζ ∂μ χ which has the right dimension. But, of course, it does not have the right Lorentz transformation because F is Lorentz-invariant. We need to include something that is dimensionless and yet carries a Lorentz index. Obviously, we want either σ μ or σ¯ μ . We know, from the Dirac equation [see Eq. (2.26)] that it is the matrices σ¯ μ that must be applied to a left-chiral spinor to yield an expression with well-deﬁned Lorentz properties. So let’s consider ζ σ¯ μ ∂μ χ Unfortunately, this still does not have the correct behavior under Lorentz transformations. From Eq. (2.26) again, we know that σ¯ μ ∂μ χ transforms like a right-chiral spinor. On the other hand, ζ is a left-chiral spinor. So we must recall how to combine a left- and right-chiral spinor to form a Lorentz invariant. The answer is that we must contract the hermitian conjugate of either spinor with the other spinor [see Eq. (2.42)]. It will be easier here to take the hermitian conjugate of ζ and contract it with σ¯ μ ∂μ χ , so we ﬁnally take δ F = K ζ † σ¯ μ ∂μ χ where K is some constant that we will determine shortly. This has the correct dimensions and Lorentz properties. By taking the dagger, we get δ F † = K ∗ (∂μ χ † )σ¯ μ ζ What we must now ﬁnd out is whether it is possible to add the term FF† without spoiling the invariance of the lagrangian under SUSY. The variation of FF† is δ(FF† ) = K ζ † σ¯ μ (∂μ χ)F † + K ∗ (∂μ χ † )σ¯ μ ζ F

(6.95)

which is not a total derivative, so we are in trouble. What we need to do is to modify the SUSY transformation of either φ or χ in such a way that the extra terms generated will cancel Eq. (6.95). Note that the variation of the spinor kinetic

Supersymmetry Demystified

130 energy is

δ(χ † i σ¯ μ ∂μ χ ) = (δχ † )i σ¯ μ ∂μ χ + χ † i σ¯ μ ∂μ (δχ)

(6.96)

which looks a lot like Eq. (6.95)! To make the similarity more apparent, let’s rewrite Eq. (6.95) in the form (after doing an integration by parts on the second term) δ(FF† ) = (K ∗ ζ F)† σ¯ μ ∂μ χ − χ † σ¯ μ ∂μ (K ∗ ζ F)

(6.97)

It is now obvious by comparing Eqs. (6.96) and (6.97) that if we take δχ = −i K ∗ ζ F + previous transformation the variation of the kinetic term of the spinor ﬁeld will cancel out the variation of FF† (modulo a total derivative). We see that the value of the constant K is left completely arbitrary. The convention is to make the new term in the transformation of χ as simple as possible, namely, −i K ∗ ζ F ≡ ζ F which ﬁxes K = −i. Our ﬁnal result is that the lagrangian L = ∂μ φ∂ μ φ † + χ † i σ¯ μ ∂μ χ + FF†

(6.98)

is invariant (up to total derivatives, as always) under the following SUSY transformations: δφ = ζ · χ δχ = −iσ μ (iσ 2 ζ ∗ ) ∂μ φ + Fζ δ F = −iζ † σ¯ μ ∂μ χ

(6.99)

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6.8 Closure of the Algebra Let’s not lose track of the original motivation for introducing an auxiliary ﬁeld, which was to ensure that the algebra closes off-shell for all the ﬁelds! What we now need to show is that for all three ﬁelds φ, χ, and F, the commutator of two SUSY transformations gives the same result, without recourse to any equation of motion. We should start by repeating the calculation that led to Eq. (6.61) for the scalar ﬁeld. Since we have modiﬁed the transformation of the spinor, there is no guarantee, at ﬁrst sight, that we will recover the same result. However, it’s easy to check that the new term present in the transformation of χ does not change the result of Eq. (6.61). Not only that, but now we do obtain the same result for all three ﬁelds without using the equations of motion; i.e., we get (see Exercise 6.7) δβ δζ X − δζ δβ X = −i ζ † σ¯ μ β − β † σ¯ μ ζ ∂μ X

(6.100)

where X stands for any of the three ﬁelds φ, χ, or F. This ﬁnally proves that the introduction of the auxiliary ﬁeld allows closure of the algebra off-shell on all the ﬁelds of the theory. And the consequence of this is that the algebra we found for the SUSY charges holds in general. EXERCISE 6.7 Prove that Eq. (6.100) is satisﬁed by all three ﬁelds φ, χ , and F.

6.9 Quiz 1. Do the supercharges commute with the momentum operator? Do they commute with the Lorentz generators? 2. What is the deﬁnition of an auxiliary ﬁeld? 3. Explain how we could have predicted that the SUSY algebra would not close off-shell before the introduction of an auxiliary ﬁeld, even before doing any calculation. 4. Without looking at the text, write down the most general transformation law of the auxiliary ﬁeld F that is linear in the ﬁelds, of the right dimension, and with the correct Lorentz property. Of course, it must contain the SUSY parameter ζ . Recall that F is of dimension 2. 5. How many supercharges did we need to introduce (count two charges related by hermitian conjugation as independent). Why did we need to introduce that many?

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CHAPTER 7

Applications of the SUSY Algebra

We have spent a considerable amount of energy working out the supersymmetry (SUSY) algebra, so it better be useful to something! Indeed, it proves to be a powerful tool for establishing general properties of supersymmetric theories without the need to ever write down any lagrangian! We will work out several such general properties in this chapter, but it seems like a good idea to ﬁrst review how some of this works in the familiar context of the Poincar´e group.

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134

7.1 Classification of States Using the Algebra: Review of the Poincar´e Group Let’s brieﬂy review some basic facts about the representations of the Poincar´e group (no SUSY here) on one-particle states. The details were worked out by Wigner in 193952 in what has become a classic paper. Here, we simply highlight some key results. Casimir operators are operators that commute with all the generators of the group. These operators are of key importance in building representations because their eigenvalues can be used to classify the representations of the group. One Casimir operator of the Poincar´e group is the squared momentum operator P μ Pμ . This obviously commutes with the momentum generators, and the proof that it commutes with the Lorentz generators as well is straightforward. The eigenvalue of Pμ P μ is m 2 , which is therefore one of the parameters used to classify one-particle states. The second Casimir operator is not as obvious. It is built out of the so-called Pauli-Lubanski operator Wμ , deﬁned as 1 W μ ≡ μνρσ Mρσ Pν 2

(7.1)

where μνρσ is the totally antisymmetric Levi-Civita symbol in four dimensions. We will use the normalization 0123 = 1 which implies 0123 = −1 With this normalization, it is clear that 0i jk = − i jk

(7.2)

where i jk is the usual three-dimensional Levi-Civita symbol. Note that we automatically have W μ Pμ = μνρσ Mρσ Pν Pμ =0

(7.3)

which is obvious because Pν Pμ is symmetric under the exchange of those two indices, whereas the Levi-Civita symbol is antisymmetric. It is not as obvious that

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135

Pμ W μ is also equal to zero (recall that Pμ and W μ are operators), but it turns out to be true, as you will show in Exercise 7.1. Therefore, [Pμ , W μ ] = 0

(7.4)

EXERCISE 7.1 Prove that Pμ W μ = 0. With some effort, one can show that Wμ W μ also commutes with all the generators of the Poincar´e group, so that’s our second Casimir operator. To uncover the physical signiﬁcation of the Pauli-Lubanski operator, consider ﬁrst massive particles. In that case, we can choose to work in the rest frame of the particle, where its four-momentum is simply

p μ = pμ = (m, 0)

(7.5)

We will be careful to distinguish the operator P μ from its eigenvalues, which we will write as p μ . For the state with four-momentum given by Eq. (7.5), we then have Pμ | p = P0 | p = m| p

(7.6)

Let us now apply the Pauli-Lubanski operator to this state: 1 W μ | p = μνρσ Mρσ Pν | p 2 1 = μ0ρσ Mρσ P0 | p 2 m = μ0ρσ Mρσ | p 2 Because of the antisymmetry of the Levi-Civita symbol, all three indices μ, ρ, and σ must be spacelike, so we conclude that W 0 | p = 0

(7.7)

Supersymmetry Demystified

136 and

m i0 jk M jk | p 2 m = − 0i jk M jk | p 2 m i jk = M jk | p 2

W i | p =

where we have used Eq. (7.2). It turns out that the operator 12 i jk M jk is simply the total angular momentum of the particle, i.e., 1 i jk M jk = L i + S i 2

(7.8)

EXERCISE 7.2 ˆ μν of the Lorentz generators acting on a The differential operator representation M spinor is deﬁned as i μν ˆ exp ω Mμν χ ≡ χ + δω χ 2 with δω χ given in Eq. (6.81). This immediately leads to ˆ μν = σμν + i(xμ ∂ν − xν ∂μ ) M = σμν + xμ Pˆ ν − xν Pˆ μ Use this to prove Eq. (7.8). HINT: You may assume that the two sides of Eq. (7.8) are applied to a one-particle state of deﬁnite momentum so that you can replace the operators Pˆ μ and Pˆ ν by the actual four-momenta pμ and pν . We therefore have W i | p = m(S i + L i )| p

(7.9)

which, in the rest frame, reduces to W i | p = m S i | p So the Casimir operator Wμ W μ acting on this state gives Wμ W μ | p = W 0 W 0 | p − W i W i | p = −m 2 S 2 | p = −m 3 s(s + 1)| p

(7.10)

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137

We now see that the second property that can be used to specify the massive representations of the Lorentz group is the spin (or total angular momentum, if we are not working in the rest frame). Let’s classify all the possible states of a massive particle at rest. For this, we need a complete set of commuting observables. In addition to the four-momentum squared (whose eigenvalue is just m 2 ) and the square of the spin, we also may use the component of the spin along the z axis, Sz , because this operator obviously commutes with Pμ P μ and with S 2 (a fact familiar from quantum mechanics). Note that W 3 = m(L z + Sz )

(7.11)

which, in the rest frame of the particle, reduces to W 3 = m Sz . Therefore, massive states are labeled in their rest frame by three quantum numbers: p, s, and sz , with P μ | p, s, sz = m| p, s, sz Wμ W μ | p, s, sz = −m 2 S 2 | p, s, sz = −m 2 s(s + 1)| p, s, sz W3 | p, s, sz = Sz | p, s, sz = sz | p, s, sz m As we all know from quantum mechanics, sz may take 2s + 1 values, ranging from −s to +s in integer steps. Now consider massless particles. In this case, we obviously cannot go to the rest frame of the particle, but we may choose to work in a frame where the fourmomentum is given by p μ = (E, 0, 0, E)

(7.12)

We will denote the corresponding state by | p0 . We then have P μ | p0 = (P 0 , P 1 , P 2 , P 3 )| p0 = (E, 0, 0, E)| p0 or P 1 | p0 = P 2 | p0 = 0 P 0 | p0 = P 3 | p0 = E| p0

(7.13)

Supersymmetry Demystified

138 We can use these results to write

P μ | p0 = (P 3 , 0, 0, P 3 )| p0

(7.14)

The reason why this is a useful thing to do will become clear in a moment. It turns out that like P μ Pμ , W μ Wμ gives zero when acting on a massless oneparticle state. The proof of this result is a bit subtle. Obviously, we cannot simply let m go to zero in Eq. (7.10) because that equation had been obtained with the assumption that we could work in the rest frame of the particle. A detailed proof can be found in Wigner’s original paper52 or in Section 2.7.2 of Ref. 28. The fact that Wμ W μ gives zero when applied to a one-particle massless state, together with Eq. (7.3), implies that when acting on such a state, the operator W μ is proportional to the four-momentum operator W μ | p0 = h P μ | p0

(7.15)

(the meaning of the constant h will be elucidated shortly). To prove Eq. (7.15), consider W μ Pμ | p0 = (W 0 P 0 − W 1 P 1 − W 2 P 2 − W 3 P 3 )| p0 = (W 0 P 0 − W 3 P 3 )| p0 = E(W 0 − W 3 )| p0 where we have used Eqs. (7.13) and (7.14). But this must be zero because of Eq. (7.3), so we conclude that W 0 | p0 = W 3 | p0

(7.16)

Using this result, we have Wμ W μ | p0 = (W 0 W 0 − W 1 W 1 − W 2 W 2 − W 3 W 3 )| p0 = −(W 1 W 1 + W 2 W 2 )| p0

(7.17)

Since W μ Wμ is zero when acting on a massless state, we obtain W 1 | p0 = W 2 | p0 = 0

(7.18)

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139

Therefore, when acting on the state, the action of the operator W μ reduces to W μ | p0 = (W 3 , 0, 0, W 3 )| p0

(7.19)

From Eq. (7.4) we know that W 3 commutes with P 3 (because it commutes with P3 ), so we may take | p0 to be a common eigenstate of these two operators. This implies that we may write W 3 | p0 = h P 3 | p0

(7.20)

Now we see why we wrote Eq. (7.14) in that form. By comparing with Eq. (7.19), we now may conclude that W μ | p0 = h P μ | p0

(7.21)

as we had set out to prove. From Eq. (7.13) we see that the action of W μ on the state is W μ | p0 = (h E, 0, 0, h E)| p0

(7.22)

To ﬁnd out what the constant h represents, we need to work out explicitly the left-hand side of Eq. (7.22). Let’s pick μ = 3. The derivation that led to Eq. (7.9) is still valid if we simply replace P0 | p = m| p with P0 | p0 = E| p0 . We then obtain W 3 | p0 = E(Sz + L z )| p0 = E sz | p0 where we have assumed that there is no orbital angular momentum. Using this in Eq. (7.22) implies that h is simply the z component of the spin, i.e., h = sz

(7.23)

Since our state has its momentum along the z direction, we may, of course, write h = s · pˆ

(7.24)

which is nothing other than the helicity of the particle. By the way, the helicity is often denoted by λ, but we prefer to use h because we will use λ to represent a left-chiral spinor that appears in supersymmetric gauge theories.

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Supersymmetry Demystified

We have obtained this result by considering μ = 3 in Eq. (7.22), but we get the same result if we pick μ = 0, as you are invited to conﬁrm in Exercise 7.3. In contrast with massive states, massless representations of the Lorentz group are therefore completely speciﬁed by only two numbers: their energy (which then speciﬁes their four-momentum) and their helicity h! The reason we don’t have 2s + 1 states, as for massive particles, lies in Eq. (7.18), which reveals that we do not have any ladder operators to change the z component of the spin. Therefore, there is only one spin state! Does this mean that massless particles can be observed with only one helicity? Well, not quite. Recall that a physical theory also must respect CPT invariance. It turns out that CPT conjugate states have opposite helicities, so a physical theory necessarily must contain two states, with helicities h and −h. The states with the intermediate values of sz that we encounter for massive particles therefore are absent for massless particles. This is the reason the photon is observed with helicities +1 and −1 or that the graviton is predicted to have helicities +2 and −2 only. We will now repeat this exercise in the context of SUSY. EXERCISE 7.3 Prove that we again obtain h = s · pˆ if we set μ = 0 in Eq. (7.22).

7.2 Effects of the Supercharges on States We will now use the SUSY algebra to determine what states are needed to build a supersymmetric theory. Of course, we have worked out one such theory explicitly, the Wess-Zumino model, but we would like to see how the results we obtained there can be generalized. We will only consider the classiﬁcation of massless supersymmetric states. These are the only states of relevance if we are interested in the minimal supersymmetric extension of the standard model because in this theory, as is the case in the standard model itself, all masses are generated through spontaneous symmetry breaking. As we saw in the previous section, massless states are identiﬁed by their fourvector and their helicity h. To emphasize this, we will change our notation from | p0 to | p, h. What we want to now is to work out the effect of the supercharges on this state. The supercharges commute with the four-momentum operator but not with the Lorentz generators, so they can change the helicity of states but not their fourmomentum. Our ﬁrst step, then, is to compute the commutator of the supercharges with W μ . Actually, we really only need the commutator with W 0 because of Eqs. (7.16) and (7.18).

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We will ﬁrst compute [Q a , W 0 ] and come back to the hermitian conjugates of the supercharges later. Using Eqs. (7.1) and (6.83), this is given by 1 [Q a , W 0 ] = 0νρσ [Q a , Mρσ Pν ] 2 1 = 0νρσ [Q a , Mρσ ] Pν 2 1 = 0νρσ (σρσ )ab Q b Pν 2 where we have used the fact that the supercharges commute with the momentum operator. We will drop the matrix indices a and b in the next few steps to alleviate the notation; we just need to keep in mind that on the right-hand side, Q is a column vector on which the Pauli matrices are acting. The only components of Pν that do not vanish when applied to a massless state are P0 and P3 . If we take ν = 0, we get zero because of the Levi-Civita symbol. This leaves us with 1 [Q, W 0 ] = 03ρσ σρσ Q P3 2 1 = 03i j σi j Q P3 2 1 = − 3i j σi j Q P3 2

(7.25)

In Exercise 7.2 we showed that i jk σ jk = σ i which allows us to write 1 1 − 3i j σi j Q P3 = − σ 3 Q P3 2 2 Reinstating the matrix indices, our ﬁnal result is therefore is 1 [Q a , W 0 ] = − (σ 3 )ab Q b P3 2

(7.26)

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142

This is also the commutator of Q a with the operator W0 because W 0 = W0 with our choice of metric. The only elements of σ 3 that are nonzero are (σ 3 )11 = −(σ 3 )22 = 1

(7.27)

Using this in Eq. (7.26) we obtain 1 [Q 1 , W0 ] = − Q 1 P3 2

(7.28)

and [Q 2 , W0 ] =

1 Q 2 P3 2

(7.29)

Now let’s ﬁnd out what this is telling us. Recall that massless states satisfy Eq. (7.21). In particular, the effect of applying W0 is simply W0 | p, h = h E| p, h

(7.30)

We’d like to ﬁnd out what is the effect of applying the supercharges Q 1 and Q 2 on | p, h. Consider ﬁrst Q 1 | p, h

(7.31)

It’s clear that this state has four-momentum pμ because the supercharges commute with the momentum operator: Pμ Q 1 | p, h = Q 1 Pμ | p, h = Q 1 pμ | p, h = pμ Q 1 | p, h Now let’s determine the helicity of Eq. (7.31). Using a trick familiar from quantum mechanics, we write W0 (Q 1 | p, h) = [W0 , Q 1 ]| p, h + Q 1 W0 | p, h 1 Q 1 E| p, h + Q 1 h E| p, h 2 1 =E h+ Q 1 | p, h 2

=

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The end result is

1 W0 (Q 1 | p, h) = E h + 2

Q 1 | p, h

which, from Eq. (7.30), tells us that 1 Q 1 | p, h = p, h + 2 Therefore the action of the supercharge Q 1 is to raise the helicity by one-half. Note that this implies that supercharges transform bosons into fermions and vice versa, as expected. From Eq. (7.29) it is clear that Q 2 has the opposite effect of Q 1 ; i.e., the supercharge Q 2 lowers the helicity by one-half: 1 Q 2 | p, h = p, h − 2 †

†

It is natural to expect that the hermitian conjugate Q 1 and Q 2 will have the opposite effect of Q 1 and Q 2 . One indeed can show that this is the case using the commutator (6.86). The calculation is almost identical to the one we followed from Eq. (7.25) to Eq. (7.29) except that it involves supercharges with upper dotted indices. Since Chapter 3 was not prerequisite material for this chapter, we will only quote the ﬁnal result, leaving the derivation to Exercise 7.4, which you are invited to do when you have become familiar with the van der Waerden notation. The end result is

1 † † Q 1 , W0 = Q 1 P3 2

† 1 † Q 2 , W0 = − Q 2 P3 2

(7.32) †

†

By comparing with Eqs. (7.28) and (7.29), we see that, as expected, Q 1 and Q 2 lower and raise the helicity by one-half, respectively. EXERCISE 7.4 Once you have become acquainted with the van der Waerden notation of Chapter 3, prove Eq. (7.32) using Eq. (6.86).

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7.3 The Massless SUSY Multiplets Now that we know the action of the supercharges on one-particle states, we will use the information to build irreducible representations of SUSY, also called SUSY multiplets or supermultiplets for short. To ﬁnd these, we will work out the minimal set of states connected by the application of the supercharges. These states will represent the particle content necessary to build supersymmetric theories. Again, we restrict ourselves to massless states. Our starting point is the SUSY algebra, Eqs. (6.70), (6.71), and (6.72). Using once again a frame in which P 1 | p, h = P 2 | p, h = 0 P 3 | p, h = P 0 | p, h = E|h

(7.33)

we have

μ

σ Pμ | p, h = P σ − P σ | p, h = 0

0

3

3

0 0

0 2P 0

| p, h

(7.34)

Recall the anticommutation relations (6.72): †

{Q a , Q b } = (σ μ )ab Pμ

(7.35)

Using this and Eq. (7.34), we conclude that †

†

†

{Q 1 , Q 1 }| p, h = {Q 1 , Q 2 }| p, h = {Q 2 , Q 1 }| p, h = 0

(7.36)

whereas †

{Q 2 , Q 2 }| p, h = 2P 0 | p, h

(7.37)

The ﬁrst relation in Eq. (7.36) is particularly useful. Its expectation value in a state | p, h is †

†

†

p, h|{Q 1 , Q 1 }| p, h = p, h|Q 1 Q 1 | p, h + p, h|Q 1 Q 1 | p, h 2 † = Q 1 | p, h + ||Q 1 | p, h||2 =0

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Since we are working with physical states whose norm is positive deﬁnite, this implies that †

Q 1 | p, h = Q 1 | p, h = 0 †

We therefore only need to concern ourselves with the supercharges Q 2 and Q 2 . What general statements can we make about the states forming a supermultiplet? They will all have the same four-momentum because the supercharges commute with the momentum operator, but they will, of course, have different helicities. It is clear that each supermultiplet has a state of minimum helicity because Q 2 is the only operator at our disposal that lowers the helicity, and Q 22 = 0. Let us then start with the state in a supermultiplet with the lowest helicity, which we will denote by | p, h min By deﬁnition, acting with Q 2 on this state gives zero. On the other hand, acting † with Q 2 gives † Q 2 | p, h min

1 = p, h min + 2 †

We cannot generate states of higher helicity because (Q 2 )2 = 0. So this is it! A massless supermultiplet contains only two states of the same momentum but with helicities differing by half a unit (and therefore, one state is a boson, whereas the second is a fermion). To be precise, our result applies to the so-called N = 1 SUSY, which involves only one Weyl spinor of charges Q (and its hermitian conjugate Q † ). One can write down a supersymmetric with any number N of spinor charges, and these are often referred to as extended supersymmetries. We will brieﬂy discuss extended supersymmetries at the end of this chapter. As far as N = 1 SUSY is concerned, we therefore need only two states to build a theory. But it would not be physical because it would not be CPT-invariant. Recall that CPT changes the sign of the helicity, so the two members of a massless supermultiplet cannot be CPT conjugate states. A physical supersymmetric theory then must contain a supermultiplet plus its CPT conjugate, for a total of four states. To be precise, here we are counting only physical, on-shell states. In the next section we will discuss in more details the massless supermultiplets. We will not discuss massive representations. The analysis is only slightly more involved, but for our purposes, only the massless representations will matter. Of course, except for the photon and the gluon, all particles are massive, but in the

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minimal supersymmetric standard model, we build the theory using massless multiplets and then incorporate their masses through the spontaneous symmetry breaking of the SU (2) L × U (1)Y group, as is done in the standard model.

7.4 Massless SUSY Multiplets and the MSSM In this section we discuss only the multiplets of N = 1 SUSY. Let us ﬁrst consider a state with h min = 0, which corresponds to a scalar ﬁeld. Then SUSY requires the presence of an h = 1/2 state as well, which is a right-handed Weyl spinor. Since for a massless spinor handedness and chirality are equivalent, this is also a right-chiral spinor. CPT invariance further requires the introduction of a second h = 0 state as well as an h = −1/2 state. The two h = 0 states can be assembled into a complex scalar ﬁeld, whereas the two fermion states correspond to a left-chiral Weyl spinor and its antiparticle. Of course, it is arbitrary which state is called the particle and which is the antiparticle, so we could just as well talk of a right-chiral particle together with its left-chiral antiparticle. In any case, the end result is that we obtain a scalar ﬁeld plus a Weyl spinor and their antiparticles. This type of multiplet is referred to as a chiral multiplet because the spin 1/2 fermion appears in only one chiral state (with the antifermion having the opposite chirality, of course). This is obviously the multiplet appearing in our model in Chapter 5. Note that the states we discuss here are the on-shell, physical states. Off shell, there are, of course, more degrees of freedom, but our discussion pertains only to physical states. In the minimal supersymmetric standard model (MSSM) all the known fermions are taken to be members of chiral multiplets. Therefore, each fermion is paired up with a spin 0 supersymmetric partner. The accepted notation is to add a preﬁx s to the names of the known fermions to designate their spin 0 partners. So each fermion is paired up with a sfermion. The quarks are paired up with squarks, and leptons are paired up with sleptons. This terminology is even used with individual particle names, the scalar partner of the electron being called the selectron, whereas the supersymmetric partner of the top quark is referred to as the stop! This whimsical notation sometimes causes problems (The scharm, anyone? What about the sstrange?). The Higgs ﬁeld is also part of a chiral multiplet, being, of course, the scalar component. Actually, the Higgs content of the MSSM is more complex than in the standard model, as we will discuss in detail in Chapter 15, and requires the introduction of two SU (2) doublets of complex Higgs ﬁelds, as opposed to the single doublet of the standard model. In any case, all these scalar ﬁelds belong to chiral multiplets and therefore have spinor superpartners. The convention for the names of spin 1/2 supersymmeric partners of standard model bosons is to add the

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sufﬁx ino at the end of the name. The spin 1/2 partners of the Higgs ﬁelds therefore are referred to as Higgsinos. Next, let’s consider supermultiplets with h min = 1/2. SUSY requires a second state with h = 1. CPT then requires states with h = −1 and h = −1/2. We can think again of the two fermion states as a left-chiral Weyl spinor together with its antiparticle. The states h = ±1 obviously correspond to the two helicity states of a massless vector boson. This type of multiplet is therefore referred as a vector multiplet or sometimes as a gauge multiplet when the vector particle is a gauge ﬁeld. All the gauge bosons of the standard model belong to vector multiplets. Following the convention described earlier, their spin 1/2 superpartners are generically called gauginos. More speciﬁcally, the superpartner of the photon, gluons, and weakinteraction bosons are, respectively, called the photino, the gluinos, the winos, and the zino. I am not making this up! For h min = 1, the four states have h = ±1, ±3/2. Again, we have the two helicity states of a massless vector particle. As for the massless spin 3/2 state, it is sometimes referred to as a Rarita-Schwinger ﬁeld. For h min = 3/2, we get four states with helicities h = ±3/2, ±2. The massless spin 2 particle is, of course, the famous graviton, and its spin 3/2 partner is called the gravitino. These are, of course, crucial elements of supergravity (SUGRA) models and string theory.

7.5 Two More Important Results We have seen how the algebra can be used to construct the massless irreducible representations of N = 1 SUSY without ever writing a lagrangian! We will now prove two additional important results using only the algebra. First, we will prove that the number of boson states must be equal to the number of fermion states in a supermultiplet. The proof will be valid for massive as well as massless representations and for any value of N . The ﬁrst step is to introduce the fermion number operator N f , which is deﬁned, as its name suggests, to give the number of fermions present in a state when it is applied to that state. Now consider the operator (−1) N f . Applied to a bosonic state, this gives +1, and applied to a fermionic state, it gives −1. If we sum the expectation value of this operator over all the states belonging to a supermultiplet, we will get for an answer the number of bosons minus the number of fermions appearing in this multiplet. Let’s carry out this calculation. First, since the supercharges change a fermion into a boson, or vice versa, we have the obvious relation (−1) N f Q a = −Q a (−1) N f

(7.38)

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Supersymmetry Demystified †

The same relation holds if we replace Q a by Q a . This implies that the product of two supercharges commutes with (−1) N f . The trick now is to sum the expectation value of †

(−1) N f {Q a , Q b } over all the states belonging to a supermultiplet. In other words, we will take the trace of this operator over the members of a supermultiplet. A few simple manipulations then give † † † Tr (−1) N f {Q a , Q b } = Tr (−1) N f Q a Q b + (−1) N f Q b Q a † † = Tr (−1) N f Q a Q b + Q b Q a (−1) N f † † = Tr −Q a (−1) N f Q b + Q b Q a (−1) N f where we have used Eq. (7.38) in the last line. Now we use the invariance of the trace under cyclic permutation of the arguments, Tr(ABC) = Tr(CAB), to obtain † † † † Tr −Q a (−1) N f Q b + Q b Q a (−1) N f = Tr −Q b Q a (−1) N f + Q b Q a (−1) N f =0 On the other hand, using the anticommutator (6.72), we also have that † Tr (−1) N f {Q a , Q b } = Tr (−1) N f σ μ Pμ Since we have proved that the left-hand side vanishes identically, independent of the four-momentum, we must have Tr(−1) N f = 0 in a SUSY supermultiplet, which implies that the number of fermions must be equal to the number of bosons. This is a general result and a well-known property of supersymmetric theories. As a second proof, we will work out a condition on the expectation value of the hamiltonian in supersymmetric theories. This leads to a result that is a hallmark of SUSY, a result that will play a signiﬁcant role in our discussion of SUSY breaking in Chapter 14. The hamiltonian is the operator P 0 . To isolate it in Eq. (6.72), we may use a simple trick. The trick becomes more obvious if we expand out the sum over Lorentz

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indices in Eq. (6.72): †

{Q a , Q b } = (σ 0 )ab H − (σ i )ab P i where σ 0 is simply the 2 × 2 identity matrix. What sets apart σ 0 from the Pauli matrices is that its trace is not zero. We therefore can isolate the hamiltonian by taking the trace of both sides, which means summing over a and b with the condition a = b. This leads to †

†

{Q 1 , Q 1 } + {Q 2 , Q 2 } = 2H Consider now calculating the expectation value of the two sides in an arbitrary state |ψ: †

†

†

†

ψ|Q 1 Q 1 |ψ + ψ|Q 1 Q 1 |ψ + ψ|Q 2 Q 2 |ψ + ψ|Q 2 Q 2 |ψ = 2ψ|H |ψ (7.39) Each term on the left-hand side is semipositive deﬁnite because it represents the norm squared of a state. For example, †

ψ|Q 1 Q 1 |ψ = Q 1 |ψ2 Since the left-hand side of Eq. (7.39) is semipositive deﬁnite, we have ψ|H |ψ ≥ 0 If we consider the special case of the state being the vacuum, then Eq. (7.39) gives the vacuum expectation value (vev) of the hamiltonian: †

†

†

†

0|Q 1 Q 1 |ψ + 0|Q 1 Q 1 |0 + 0|Q 2 Q 2 |0 + 0|Q 2 Q 2 |0 = 20|H |0

(7.40)

and we ﬁnd again that 0|H |0 ≥ 0

(7.41)

with the inequality saturated if the supercharges annihilate the vacuum. If they do not annihilate the vacuum, then we have a spontaneous breaking of SUSY, and we get a strict inequality, i.e., 0|H |0 > 0

(7.42)

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150

Therefore, a necessary and sufﬁcient condition for SUSY to be spontaneously broken is that the vacuum expectation value of the hamiltonian is nonzero (and necessarily positive). We will come back to this issue in Chapter 14.

7.6 The Algebra in Majorana Form Many references give the SUSY algebra in Majorana form instead of the Weyl form we just worked out. We will now show how to translate all the commutation relations involving the supercharges into the Majorana language. The ﬁrst step is to group the supercharges into a Majorana spinor. Recall from Eq. (4.20) that a Majorana spinor can be built out of a left-chiral Weyl spinor χ and the hermitian conjugate χ † via

M =

iσ 2 χ †T χ

We therefore can assemble the supercharges into a Majorana supercharge QM ≡

iσ 2 Q †T Q

which, written explicitly, is † ⎞ Q2 ⎜ −Q † ⎟ 1⎟ =⎜ ⎝ Q1 ⎠ Q2

⎛

QM

(7.43)

Let us deﬁne the components Q M1 , Q M2 , . . . through ⎞ Q M1 ⎜ Q M2 ⎟ ⎟ ≡⎜ ⎝ Q M3 ⎠ Q M4 ⎛

QM

Then all the anticommutators among the supercharges can be shown to be reproduced by the single expression ¯ Mb } = (γ μ )ab Pμ {Q Ma , Q

(7.44)

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¯ Mb stands for Here, the bar over Q M represents the usual Dirac adjoint, so Q † 0 (Q M γ )b . If we deﬁne the matrices γμν by i γμν ≡ (γμ γν − γν γμ ) 4 then the commutation relation of the supercharges with the Lorentz generators are [Q Ma , Mμν ] = (γμν )ab Q Mb It is not difﬁcult to show that this expression reproduces both Eqs. (6.83) and (6.86) using the fact that γμν may be written as γμν =

σ¯ μν

0

0

σμν

(7.45)

We will now obtain the SUSY charges from the supersymmetric currents, following Eq. (6.24). But ﬁrst let’s review the general approach. EXERCISE 7.5 Verify that Eq. (7.44) reproduces the anticommutators (6.70), (6.71), and (6.72).

7.7 Obtaining the Charges from Symmetry Currents To see explicitly how to construct the currents, consider the theory of a single complex scalar ﬁeld. If the ﬁeld is varied by an amount δφ, the lagrangian varies by δL(φ, φ † ) =

∂L ∂L ∂L ∂L δφ + δ(∂μ φ) + † δφ † + δ(∂μ φ † ) ∂φ ∂(∂μ φ) ∂φ ∂(∂μ φ † )

This is simply an application of the chain rule from ordinary multivariable calculus (a similar trick is used to determine the equations of motion of the ﬁelds). If this transformation is a symmetry of the theory, it leaves the action invariant, which, as we have discussed previously, implies either that the lagrangian density

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152

is invariant or that it changes by a total derivative. To be general, let’s assume that it changes by a total derivative, which we will write as ∂μ K μ , with K μ some function of the ﬁelds. We then have ∂L ∂L ∂L ∂L δφ + ∂μ (δφ) + † δφ † + ∂μ (δφ † ) = ∂μ K μ ∂φ ∂(∂μ φ) ∂φ ∂(∂μ φ † )

(7.46)

where we have used δ(∂μ φ) = ∂μ (δφ) and δ(∂μ φ † ) = ∂μ (δφ † ). Now, assuming that φ and φ † obey the equations of motion, we have ∂L ∂L = ∂μ ∂φ ∂(∂μ φ)

(7.47)

∂L ∂L = ∂μ ∂φ † ∂(∂μ φ † )

(7.48)

and

Substituting Eqs. (7.47) and (7.48) into Eq. (7.46), we get ∂μ

∂L ∂L † δφ + δφ = ∂μ K μ ∂(∂μ φ) ∂(∂μ φ † )

(7.49)

The quantity in brackets is called the Noether current j μ , i.e., jμ ≡

∂L ∂L δφ + δφ † ∂(∂μ φ) ∂(∂μ φ † )

so Eq. (7.49) may be written as ∂μ j μ = ∂μ K μ Therefore, the quantity J μ ≡ ( j μ − K μ) is conserved, in the sense that it satisﬁes ∂μ J μ = 0

(7.50)

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Of course, in a theory where there are several ﬁelds, such as the supersymmetric theory we built in Chapters 5 and 6, we have to add terms such as Eqs. (7.47) and (7.48) for each ﬁeld. We will illustrate this later in this chapter. Now we get to the key point. The claim is that the charges generating the transformations are given by the spatial integral of the zeroth component of the currents: Q=

d3x J 0

(7.51)

We won’t prove that the charges are indeed given in general by Eq. (7.51) (see any textbook on quantum ﬁeld theory for a derivation). As an aside, the Q are often described as conserved charges because they are time-independent. As already emphasized, it is important to keep in mind that the currents, and hence the charges constructed via the currents, are functions of the quantum ﬁelds appearing in the theory, so they are themselves quantum ﬁeld operators. Once the charges are found using Eq. (7.51), their commutation (or anticommutation) relations can be worked out explicitly using the equal time commutation/anticommutation of the ﬁelds. Let’s now apply this to SUSY!

7.8 Explicit Supercharges as Quantum Field Operators Now let us build the SUSY current and ﬁnd explicit expressions for the supercharges in terms of the ﬁelds of the theory. In the case of SUSY, Eq. (7.51) is replaced by ¯ · ζ¯ = Q·ζ + Q

0 d 3 xJSUSY

We work with the free supersymmetric lagrangian, which we will repeat here: L = ∂μ φ∂ μ φ † + χ † i σ¯ μ ∂μ χ + FF †

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First, let’s ﬁnd the SUSY Noether currents. In order to do this, we need ∂L δφ = (∂ μ φ † )δφ ∂(∂μ φ) = (∂μ φ † )χ · ζ ∂L δχ = χ † i σ¯ μ δχ ∂(∂μ χ ) = χ † σ¯ μ (∂ν φ)σ ν iσ 2 ζ ∗

(7.52) (7.53) (7.54) (7.55)

∂L δφ † = (∂ μ φ)χ¯ · ζ¯ ∂(∂μ φ † ) On the other hand, ∂L ∂L ∂L = = =0 † ∂(∂μ F) ∂(∂μ F ) ∂(∂μ χ † ) The Noether current is therefore j μ = (∂μ φ † )χ · ζ + (∂ν φ)χ † σ¯ μ σ ν iσ 2 ζ ∗ + (∂ μ φ)χ¯ · ζ¯ To ﬁnd ∂μ K μ , we simply have to add up the total derivatives we dropped when we calculated the supersymmetric transformation of our lagrangian. These total derivatives can be found by comparing Eqs. (5.14) and (5.15) to the equations preceding them, giving us (recall that we showed that the constant C appearing in those equations was equal to −i) ∂μ K μ = ∂μ χ¯ · ζ¯ ∂ μ φ − i(∂ν φ † )ζ T iσ 2 σ ν i σ¯ μ χ + (∂ μ φ † )ζ · χ

(7.56)

where we have used χ † iσ 2 ζ ∗ = χ¯ · ζ¯ and ζ T iσ 2 χ = −ζ · χ. So K μ is simply the expression inside the parenthesis. Using the fact that the order in spinor dot products does not matter, the supersymmetric current is simply μ

JSUSY = j μ − K μ = (∂ν φ)χ † σ¯ μ σ ν iσ 2 ζ ∗ − (∂ν φ † )ζ T iσ 2 σ ν σ¯ μ χ

(7.57)

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We can then use this to ﬁnd the charges: ¯ · ζ¯ = ζ T (−iσ 2 )Q + Q † iσ 2 ζ ∗ ζ·Q+Q = d 3 x (∂ν φ)χ † σ¯ 0 σ ν iσ 2 ζ ∗ − (∂ν φ † )ζ T iσ 2 σ ν σ¯ 0 χ ¯ as Q ¯ · ζ¯ . This was simply in order to On the left-hand side we chose to write ζ¯ · Q get ζ ∗ to appear on the right of the σ 2 matrix, as in the current. The matrix σ¯ 0 is simply the identity matrix. Since the inﬁnitesimal parameters ζ and ζ ∗ may be treated as independent, we can read off the charges Q and Q † . We ﬁnd Q = d 3 x(∂ν φ † (x))σ ν χ (x) (7.58) and †

Q =

d 3 xχ † (x)∂ν φ(x)σ ν

(7.59)

which is simply the hermitian conjugate of Q. This is what we were looking for: explicit expressions for the supercharges as quantum ﬁeld operators. It is then easy to verify explicitly that these charges indeed generate the SUSY transformations of the ﬁelds given in Eq. (6.52). This will be left as an exercise. All we will need are the equal time commutators and anticommutators familiar from quantum ﬁeld theory:

φ( x , t), φ˙ † ( y , t) = iδ 3 ( x − y ) †

{χa ( x , t), χb ( y , t)} = δab δ 3 ( x − y )

(7.60)

where a dot indicates a derivative with respect to time. All other commutators and ˙ = 0. anticommutators are equal to zero. For example, [φ, φ] The commutators of the supercharges with the Poincar´e generators also can be veriﬁed, but this requires that we ﬁrst work out Pμ and Mμν as quantum ﬁeld operators. We will not pursue this here because, although straightforward, the calculation is a bit lengthy and is in any case similar to usual quantum ﬁeld theory calculations. The interested reader will ﬁnd the detailed calculation in Ref. 31. EXERCISE 7.6 Obtain the commutators (6.52) using the quantum ﬁeld operator representations of the supercharges (7.58) and (7.59) and the equal time commutators (7.60).

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7.9 The Coleman-Mandula No-Go Theorem A book on SUSY would be incomplete if it did not mention one of the most famous no-go theorems of particle physics. Now that we have worked a bit with charges and their algebra, the theorem will be easier to understand. Since the discovery of relativity, it has been clear that particle states should belong to representations of the Poincar´e group. On the other hand, it is natural to also group particles into multiplets of some additional “internal” symmetry group, and this viewpoint, of course, has been very successful in the construction of gauge theories. An obvious question to ask is whether such a symmetry group actually could “mix” nontrivially with the Poincar´e group. To make this rather vague statement more precise, the question is whether the generators of the internal symmetry group could have nonzero commutators with the generators of the Poincar´e group. Let us write the charges (or generators) of the internal symmetry group as T a and their algebra as [T a , T b ] = i f abc T c

(7.61)

The question is whether it is possible to write down some physically meaningful theory where the charges T a do not commute with the generators of the Poincar´e group Mμν and Pμ (another way to put it is to say that the T a should carry some Lorentz index). This is an interesting question because such a theory then would mix spacetime transformations with “internal” transformations and would lead to a highly nontrivial extension of gauge theories. What Coleman and Mandula proved in 196711 (a derivation also can be found 47 in ) was that for an algebra of the form of Eq. (7.61) and under a few very reasonable conditions, e.g., that elastic scattering amplitudes must be analytic functions of the Mandelstam variables and that plane waves must in general scatter, the answer is a resounding no! In other words, the generators T a cannot carry spacetime indices. Exercise 7.7 illustrates the theorem in the simple case of a symmetric second-rank tensor charge Sμν = Sνμ . EXERCISE 7.7 Consider a symmetric tensor Sμν that is conserved. Then its eigenvalue will be conserved in a collision. Consider the scattering process a + b → a + b. Denote the four-momenta of the particles pia , pib , p af , and p bf . Show that the conservation of the eigenvalue of Sμν together with the conservation of total momentum implies that either there is no scattering at all, p af = pia and p bf = pib , or that the particles exchange their momenta, paf = pib and p bf = pia . No other direction of scattering is possible! This is clearly unphysical.

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Hint: Write the most general eigenvalue for this operator for a one-particle state. Then use that the corresponding eigenvalue for a two-particle state is simply the sum of the one-particle state eigenvalues. This seemed to be the kiss of death for the idea of building theories mixing spacetime transformations and internal symmetries nontrivially. One might think that physicists would have been eager to ﬁnd some way to circumvent this no-go theorem as soon as it came out, but this was not how SUSY was ﬁrst discovered. We won’t get into the history of SUSY here; interested readers are invited to read the ﬁrst chapter of Ref. 47 for a very informative historical summary. Of course, SUSY does circumvent the theorem in a very simple way: The supercharges do not obey a set of commutation relations as in Eq. (7.61) but instead obey anticommutation relations. Another way to carry the same information is that supercharges transform as a spinor representation of the Lorentz group instead of transforming as vector (or more generally, tensor) representations. The theorem is still useful because it tells us that no other symmetry with charges obeying commutation rules could mix nontrivially with spacetime transformations. But are the supercharge anticommutators we wrote down the most general possible ones? The next section addresses this question.

7.10 The Haag-Lopuszanski-Sohnius Theorem and Extended SUSY The discovery of SUSY raised another obvious question: Is the algebra we wrote down the most general algebra that evades Coleman and Mandula’s theorem? The answer was provided by Haag, Lopuszanski, and Sohnius in 1975,23 and the answer again was a resounding no. Not only did they show that other possibilities exist, but they also worked out the most general algebra possible. To write it down, we ﬁrst introduce a set of N copies of the supercharges, i.e., †I

†I

Q 1I , Q 2I , Q 1 , Q 2

where I is an extra label ranging from 1 to N . The SUSY we have considered so far in this book corresponds to N = 1 and therefore is referred to as N = 1 supersymmetry. The most general algebra satisﬁed by these charges is I Q a , Q bJ = (−iσ 2 )ab Z I J I J † Q a , Q b = (σ μ )ab δ I J Pμ I † J † Q a , Q b = (−iσ 2 )ab (Z I J )∗ (7.62)

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where the Z I J are called the central charges of the algebra (a central charge refers to a quantity which commutes with all the generators of an algebra) and always can be made antisymmetric, Z I J = −Z J I . In the case of N = 1 supersymmetry, the unique central charge Z 11 is therefore necessarily zero, and the algebra Eq. (7.62) reduces to what we obtained earlier in this chapter. Extended SUSY is beyond the scope of this book, but one important point to make is that from a purely phenomenological point of view, only N = 1 is attractive, at least in theories describing physics at the low energies that will be accessible to the Large Hadron Collider (LHC). The reason is that starting with N = 2, supersymmetric theories are necessarily nonchiral; i.e., they must treat left- and right-chiral states on equal footing. For N = 2, the chiral supermultiplet contains both a rightand a left-chiral fermion, with h = 1/2 and h = −1/2 (as well as two states with h = 0). But the members of a supermultiplet all transform the same way under a gauge symmetry, so for N = 2 (and any larger value of N as well), the two chiral states must be treated on an equal footing by any gauge force. But this conﬂicts with what we know about weak force of the standard model, which treats differently left- and right-chiral states; hence only N = 1 SUSY is relevant to phenomenology. This is true if we think of the MSSM as some effective ﬁeld theory to be used with energies on the order of a few TeV. A more fundamental theory could have a very different structure, and who knows how symmetry may be broken down to what we observe at small energies relative to the Planck scale. But even if we set aside the issue of chirality, other problems arise, if we increase too much the value of N because it does not seem possible to write down consistent quantum ﬁeld theories containing ﬁelds with spin larger than 2. More details on extended SUSY may be found in Refs. 8 and 47.

7.11 Quiz 1. What is the effect of the supercharge Q 1 on the four-momentum and helicity of a state? 2. For N = 1, how many states are required to form a SUSY multiplet? 3. What is extended SUSY? 4. What can we say about the vacuum expectation value of the hamiltonian in a supersymmetric theory? 5. How does SUSY evade the Coleman-Mandula no-go theorem?

CHAPTER 8

Adding Interactions: The Wess-Zumino Model

So far we have considered a rather uninteresting theory: All the ﬁelds are massless, and even more boring, there are no interactions at all. What we would like to do now is to see if we can include masses and interactions that will preserve supersymmetry (SUSY). The answer is obviously yes because SUSY would never have become such a hot topic if interactions could not be included. However, as you surely expect, this will prove to be a highly nontrivial undertaking. In writing down possible interaction terms, we will follow certain rules:

• We will not consider terms of dimension above 4. This requires some explanation. Of course, any term in a lagrangian density is of dimension 4. What we mean here is that if we ignore all the constants, the dimension must not be more than 4. Consider for example the four-fermion interaction χ · χ χ¯ · χ¯ . Since the dimension of fermion ﬁelds is 3/2, this term has dimension 6 and must be

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rejected. Note that if we insisted on including this term in our lagrangian density, we would have to divide it by a constant having the dimension of energy squared: Lint =

1 χ · χ χ¯ · χ¯ μ2

(8.1)

where μ would be a free parameter with dimension of energy. Therefore, another way of saying that we reject terms of dimension more than 4 is that we will not consider any term in the lagrangian that contains a dimensionful constant in the denominator. The reason for this condition is that any interaction of dimension 5 or higher leads to a nonrenormalizable theory. There is nothing intrinsically wrong with nonrenormalizable theories if they are treated as effective ﬁeld theories, but we won’t consider these in this book. • All the terms we add must, of course, be Lorentz-invariant. This obviously implies that any upper (contravariant) Lorentz index must be contracted with a corresponding lower (covariant) Lorentz index. But this is not all. We also must ensure that spinors are combined into Lorentz-invariant combinations. This is where all our work of Chapter 2 will come in handy. • The lagrangian must be real, L† = L, in order to have CPT invariance. • And ﬁnally, of course, we will insist that the lagrangian is invariant (up to total derivatives) under the SUSY transformations of the ﬁelds, which we repeat here for convenience: δφ = ζ · χ = ζ T (−iσ 2 )χ δφ † = χ¯ · ζ¯ = χ † iσ 2 ζ ∗ δχ = −iσ μ (∂μ φ)iσ 2 ζ ∗ + Fζ δ F = −iζ † σ¯ μ ∂μ χ

(8.2)

8.1 A Supersymmetric Lagrangian With Masses and Interactions Let’s begin. We want to write new terms built out of the ﬁelds φ, χ, F, and their hermitian conjugates that will respect our four criteria. Consider ﬁrst interactions that depend only on φ and its dagger φ † . To keep things general, let’s just write this

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interaction as L1 = G(φ, φ † )

(8.3)

This is automatically Lorentz-invariant because φ is a scalar ﬁeld. We use a calligraphic G instead of F for this function in order to avoid possible confusion with the auxiliary ﬁeld F. To satisfy the criterion of renormalizability, we may only consider functions of at most four scalar ﬁelds. Without loss of generality, we may take G to be real, G † = G (if it weren’t, we would simply add to it its hermitian conjugate and rename the sum G). What else can we write down? Well, we can consider some arbitrary function of φ times the auxiliary ﬁeld F. We will write this term as W1 (φ, φ † )F

(8.4)

The notation W1 is a standard one. The reason for the subscript 1 is that we are dealing with a single set of ﬁelds φ, χ , and F. Later on we will extend our discussion to consider several sets of ﬁelds, and the 1 will be replaced by an index i. The lagrangian Eq. (8.4) is automatically Lorentz-invariant because both φ and F are scalar ﬁelds. For this to be of dimension 5 or less, W must contain the product of at most two φ or φ † ﬁelds. However, Eq. (8.4) is not real. To get a real lagrangian, we must add to Eq. (8.4) its hermitian conjugate, so we will consider instead †

L2 = W1 F + W1 F †

(8.5)

Note that there is no reality condition imposed on W1 . Next, we also can consider a general function of the scalar ﬁelds multiplying a Lorentz-invariant combination of the spinor ﬁeld χ . We have seen that a possible Lorentz invariant made of χ is χ · χ, so we write this new contribution to the lagrangian as 1 − W11 (φ, φ † )χ · χ 2

(8.6)

Once again, we are using a standard notation for the name of the function of the scalar ﬁeld, and the factor of −1/2 is purely conventional and chosen to simplify a later result. There is no a priori relation between W11 and W1 , although we will later ﬁnd that they must indeed be closely related in order for the theory to be supersymmetric.

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Again, Eq. (8.6) is not real, so we need to consider instead the real combination [using Eq. (A.44)]: 1 1 † L3 = − W11 χ · χ − W11 (χ · χ )† 2 2 1 1 † = − W11 χ · χ − W11 χ¯ · χ¯ 2 2

(8.7)

There is no reality condition imposed on W11 . Next, we might think about considering terms containing a function of the auxiliary ﬁeld F times χ · χ , but this is necessarily of dimension 5 or higher. For the same reason, it is impossible to write any term that contains φ (or φ † ) and two auxiliary ﬁelds or to write terms containing all three types of ﬁelds (these start at dimension 6). Denoting by L0 the free supersymmetric lagrangian obtained in Chapter 6, the most general lagrangian that is real, renormalizable, and Lorentz-invariant is therefore Ltotal = L0 + L1 + L2 + L3 †

= ∂μ φ∂ μ φ † + χ † i σ¯ μ ∂μ χ + F F † + G + W1 F + W1 F † 1 1 † − W11 χ · χ − W11 χ¯ · χ¯ 2 2

(8.8)

where one must keep in mind that G, W1 , and W11 are functions of φ and φ † . We did not include any coefﬁcients in front of the new terms because they can be absorbed into a redeﬁnition of G, W1 , and W11 . Our goal now is to see if we can ﬁnd expressions for those three functions of the scalar ﬁelds such that the total lagrangian is supersymmetric. We already know that L0 is invariant under SUSY so we only need to impose that L1 + L2 + L3 must be supersymmetric. From now on we will call the sum of these three terms Lint because they contain interactions between the different ﬁelds, i.e., L1 + L2 + L3 ≡ Lint

(8.9)

Let us write the variation of Lint under SUSY explicitly: †

δLint = δG + δ(W1 F) + δ(W1 F † ) −

1 1 † δ(W11 χ · χ) − δ(W11 χ¯ · χ¯ ) 2 2

(8.10)

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At ﬁrst sight, it may seem like we will need to write down explicit expressions for W1 , W11 , and G before we can calculate the supersymmetric variation of our lagrangian, but this won’t be necessary. For example, we can simply write the total variation of G(φ, φ † ) under a SUSY transformation as δG =

∂G ∂G δφ + † δφ † ∂φ ∂φ

(8.11)

using the rule familiar from calculus. We therefore get, using Eq. (8.2), δG =

∂G ∂G ζ · χ + † χ¯ · ζ¯ ∂φ ∂φ

(8.12)

We can, of course, use the same trick to write down the variation of W1 and W11 . For example, we have δ(W1 F) = (δW1 ) F + W1 δ F ∂ W1 ∂ W1 † = δφ F + W1 δ F δφ + ∂φ ∂φ † =

∂